func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

func countOnes(str: String) -> Int {
    var count = 0
    for i in 0..<str.count {
        if str[str.index(str.startIndex, offsetBy: i)] == "1" {
            count += 1
        }
    }
    return count
}

func countZeros(str: String) -> Int {
    var count = 0
    for i in 0..<str.count {
        if str[str.index(str.startIndex, offsetBy: i)] == "0" {
            count += 1
        }
    }
    return count
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

// Input is a string of 1s and 0s.
// Return the number of 1s in the string.
// countOnes("101101") == 3
func countOnes(str: String) -> Int {
    var count = 0
    for i in 0..<str.count {
        if str[str.index(str.startIndex, offsetBy: i)] == "1" {
            count += 1
        }
    }
    return count
}

// Input is a string of 1s and 0s.
// Return the number of 0s in the string.
// countZeros("101101") == 2
func countZeros(str: String) -> Int {
    var count = 0
    for i in 0..<str.count {
        if str[str.index(str.startIndex, offsetBy: i)] == "0" {
            count += 1
        }
    }
    return count
}

// Input is a string of 1s and 0s.
// Return the number of 1s in the string.
// countZeros("101101") == 2
func countZeros(str: String) -> Int {
    var count = 0
    for i in 0..<str.count {
        if str[str.index(str.startIndex, offsetBy: i)] == "0" {
            count += 1
        }
    }
    return count
}

// Input is a string of 1s and 0s.
// Return the number of 1s in the string.
// countZeros("101101") == 2
func countZeros(str: String) -> Int {
    var count = 0
    for i in 0..<str.count {
        if str[str.index(str.startIndex, offsetBy: i)] == "0" {
            count += 1
        }
    }
    return count
}

// Input is a string of 1s and 0s.
// Return the number of 1s in the string.
// countZeros("101101") == 2
func countZeros(str: String) -> Int {
    var count = 0
    for i in 0..<str.count {
        if str[str.index(str.startIndex, offsetBy: i)] == "0" {
            count += 1
        }
    }
    return count
}

// Input is a string of 1s and 0s.
// Return the number of 1s in the string.
// countZeros("101101") == 2
func countZeros(str: String) -> Int {
    var count = 0
    for i in 0..<str.count {
        if str[str.index(str.startIndex, offsetBy: i)] == "0" {
            count += 1
        }
    }
    return count
}

// Input is a string of 1s and 0s.
// Return the number of 1s in the string.
// countZeros("101101") == 2
func countZeros(str: String) -> Int {
    var count = 0
    for i in 0..<str.count {
        if str[str.index(str.startIndex, offsetBy: i)] == "0" {
            count += 1
        }
    }
    return count
}

// Input is a string of 1s and 0s.
// Return the number of 1s in the string.
// countZeros("101101") == 2
func countZeros(str: String) -> Int {
    var count = 0
    for i in 0..<str.count {
        if str[str.index(str.startIndex, offsetBy: i)] == "0" {
            count += 1
        }
    }
    return count
}

// Input is a string of 1s and 0s.
// Return the number

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] == b[i] ? "0" : "1")
    }
    return result
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] == b[i] ? "0" : "1")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] == b[i] ? "0" : "1")
    }
    return result
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] == b[i] ? "0" : "1")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[a.index(a.startIndex, offsetBy: i)]))!
        let bBit = Int(String(b[b.index(b.startIndex, offsetBy: i)]))!
        
        result.append(String(aBit ^ bBit))
    }
    
    return result
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[a.index(a.startIndex, offsetBy: i)]))!
        let bBit = Int(String(b[b.index(b.startIndex, offsetBy: i)]))!
        
        result.append(String(aBit ^ bBit))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[a.index(a.startIndex, offsetBy: i)]))!
        let bBit = Int(String(b[b.index(b.startIndex, offsetBy: i)]))!
        
        result += String(aBit ^ bBit)
    }
    
    return result
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[a.index(a.startIndex, offsetBy: i)]))!
        let bBit = Int(String(b[b.index(b.startIndex, offsetBy: i)]))!
        
        result += String(aBit ^ bBit)
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] == b[i] ? "0" : "1")
    }
    return result
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] == b[i] ? "0" : "1")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[a.index(a.startIndex, offsetBy: i)]))!
        let bBit = Int(String(b[b.index(b.startIndex, offsetBy: i)]))!
        
        result += String(aBit ^ bBit)
    }
    
    return result
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[a.index(a.startIndex, offsetBy: i)]))!
        let bBit = Int(String(b[b.index(b.startIndex, offsetBy: i)]))!
        
        result += String(aBit ^ bBit)
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[i]))!
        let bBit = Int(String(b[i]))!
        
        result.append(String(aBit ^ bBit))
    }
    
    return result
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[i]))!
        let bBit = Int(String(b[i]))!
        
        result.append(String(aBit ^ bBit))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[a.index(a.startIndex, offsetBy: i)]))!
        let bBit = Int(String(b[b.index(b.startIndex, offsetBy: i)]))!
        
        result.append(String(aBit ^ bBit))
    }
    
    return result
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[a.index(a.startIndex, offsetBy: i)]))!
        let bBit = Int(String(b[b.index(b.startIndex, offsetBy: i)]))!
        
        result.append(String(aBit ^ bBit))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[a.index(a.startIndex, offsetBy: i)]))!
        let bBit = Int(String(b[b.index(b.startIndex, offsetBy: i)]))!
        result += String(aBit ^ bBit)
    }
    
    return result
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = Int(String(a[a.index(a.startIndex, offsetBy: i)]))!
        let bBit = Int(String(b[b.index(b.startIndex, offsetBy: i)]))!
        result += String(aBit ^ bBit)
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result += String(Int(a[i])! ^ Int(b[i])!)
    }
    return result
}

<bos><start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result += String(Int(a[i])! ^ Int(b[i])!)
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    var a = a
    var b = b
    while a.count > 0 || b.count > 0 {
        let a = a.prefix(1)
        let b = b.prefix(1)
        if a == b {
            result = "0" + result
        } else {
            result = "1" + result
        }
        a = String(a.dropFirst())
        b = String(b.dropFirst())
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    var a = a
    var b = b
    while a.count > 0 || b.count > 0 {
        let a = a.prefix(1)
        let b = b.prefix(1)
        if a == b {
            result = "0" + result
        } else {
            result = "1" + result
        }
        a = String(a.dropFirst())
        b = String(b.dropFirst())
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        result += aChar == bChar? "0" : "1"
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        result += aChar == bChar? "0" : "1"
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    return ""
}
```

func stringXor(a: String, b: String) -> String {
    return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    return ""
}
```

func stringXor(a: String, b: String) -> String {
    return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    return ""
}
```

func stringXor(a: String, b: String) -> String {
    return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    return ""
}
```

func stringXor(a: String, b: String) -> String {
    return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    return ""
}
```

func stringXor(a: String, b: String) -> String {
    return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    return ""
}
```

func stringXor(a: String, b: String) -> String {
    return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    return ""
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    var aIndex = a.startIndex
    var bIndex = b.startIndex
    while aIndex < a.endIndex && bIndex < b.endIndex {
        let aChar = a[aIndex]
        let bChar = b[bIndex]
        if aChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
        aIndex = a.index(after: aIndex)
        bIndex = b.index(after: bIndex)
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    var aIndex = a.startIndex
    var bIndex = b.startIndex
    while aIndex < a.endIndex && bIndex < b.endIndex {
        let aChar = a[aIndex]
        let bChar = b[bIndex]
        if aChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
        aIndex = a.index(after: aIndex)
        bIndex = b.index(after: bIndex)
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[i]
        let charB = b[i]
        if charA == charB {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[i]
        let charB = b[i]
        if charA == charB {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        let aChar = a.count > i? a[a.index(a.startIndex, offsetBy: i)] : "0"
        let bChar = b.count > i? b[b.index(b.startIndex, offsetBy: i)] : "0"
        if aChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        let aChar = a.count > i? a[a.index(a.startIndex, offsetBy: i)] : "0"
        let bChar = b.count > i? b[b.index(b.startIndex, offsetBy: i)] : "0"
        if aChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[i]
        let charB = b[i]
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[i]
        let charB = b[i]
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        let aChar = i < a.count? a[a.index(a.startIndex, offsetBy: i)] : "0"
        let bChar = i < b.count? b[b.index(b.startIndex, offsetBy: i)] : "0"
        let xor = (aChar == "1"? 1 : 0) ^ (bChar == "1"? 1 : 0)
        result.append(xor == 1? "1" : "0")
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        let aChar = i < a.count? a[a.index(a.startIndex, offsetBy: i)] : "0"
        let bChar = i < b.count? b[b.index(b.startIndex, offsetBy: i)] : "0"
        let xor = (aChar == "1"? 1 : 0) ^ (bChar == "1"? 1 : 0)
        result.append(xor == 1? "1" : "0")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        let aChar = a.index(a.startIndex, offsetBy: i)
        let bChar = b.index(b.startIndex, offsetBy: i)
        let aValue = a[aChar]
        let bValue = b[bChar]
        let xor = (aValue == bValue)? "0" : "1"
        result.append(xor)
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        let aChar = a.index(a.startIndex, offsetBy: i)
        let bChar = b.index(b.startIndex, offsetBy: i)
        let aValue = a[aChar]
        let bValue = b[bChar]
        let xor = (aValue == bValue)? "0" : "1"
        result.append(xor)
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        let charB = b.index(b.startIndex, offsetBy: index)
        let charB = b[charB]
        if char == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        let charB = b.index(b.startIndex, offsetBy: index)
        let charB = b[charB]
        if char == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == b[index] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == b[index] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[i]
        let charB = b[i]
        let xor = charA == charB? "0" : "1"
        result.append(xor)
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[i]
        let charB = b[i]
        let xor = charA == charB? "0" : "1"
        result.append(xor)
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[i]
        let charB = b[i]
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[i]
        let charB = b[i]
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.characters.enumerate() {
        let bChar = b[b.startIndex.advancedBy(index)]
        if char == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.characters.enumerate() {
        let bChar = b[b.startIndex.advancedBy(index)]
        if char == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.characters.enumerate() {
        let bChar = b.characters[b.startIndex.advancedBy(index)]
        if char == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.characters.enumerate() {
        let bChar = b.characters[b.startIndex.advancedBy(index)]
        if char == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    // Your code here
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Your code here
}
```

func stringXor(a: String, b: String) -> String {
    // Your code here
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Your code here
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    // Your code here
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Your code here
}
```

func stringXor(a: String, b: String) -> String {
    // Your code here
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Your code here
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.characters.enumerate() {
        let bChar = b.characters[b.startIndex.advancedBy(index)]
        let xor = char == bChar? "0" : "1"
        result.append(xor)
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.characters.enumerate() {
        let bChar = b.characters[b.startIndex.advancedBy(index)]
        let xor = char == bChar? "0" : "1"
        result.append(xor)
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    // Your code here
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Your code here
}
```

func stringXor(a: String, b: String) -> String {
  return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
  return ""
}
```

func stringXor(a: String, b: String) -> String {
    let aChars = a.characters.map { String(format: "%04b", $0.value) }
    let bChars = b.characters.map { String(format: "%04b", $0.value) }
    let resultChars = aChars.enumerated().map { (index, aChar) in
        let bChar = bChars[index]
        let result = aChar.binaryXOR(bChar)
        return String(format: "%04b", result)
    }
    return String(resultChars)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aChars = a.characters.map { String(format: "%04b", $0.value) }
    let bChars = b.characters.map { String(format: "%04b", $0.value) }
    let resultChars = aChars.enumerated().map { (index, aChar) in
        let bChar = bChars[index]
        let result = aChar.binaryXOR(bChar)
        return String(format: "%04b", result)
    }
    return String(resultChars)
}
```

func stringXor(a: String, b: String) -> String {
    let aAsBits = a.map { String(format: "%04b", $0) }
    let bAsBits = b.map { String(format: "%04b", $0) }
    
    let result = aAsBits.enumerated().map { index, bit in
        let aBit = bit == "1"? 1 : 0
        let bBit = bAsBits[index] == "1"? 1 : 0
        return (aBit ^ bBit).description
    }
    
    return result.joined()
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aAsBits = a.map { String(format: "%04b", $0) }
    let bAsBits = b.map { String(format: "%04b", $0) }
    
    let result = aAsBits.enumerated().map { index, bit in
        let aBit = bit == "1"? 1 : 0
        let bBit = bAsBits[index] == "1"? 1 : 0
        return (aBit ^ bBit).description
    }
    
    return result.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aAsBits = a.map { String(format: "%04b", $0) }
    let bAsBits = b.map { String(format: "%04b", $0) }
    let result = aAsBits.enumerated().map { index, bit in
        bit == "1"? bAsBits[index] == "1"? "0" : "1" : bAsBits[index]
    }
    return result.joined()
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aAsBits = a.map { String(format: "%04b", $0) }
    let bAsBits = b.map { String(format: "%04b", $0) }
    let result = aAsBits.enumerated().map { index, bit in
        bit == "1"? bAsBits[index] == "1"? "0" : "1" : bAsBits[index]
    }
    return result.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aInt = a.reduce(0) { $0 * 2 + Int(String($1), radix: 2)! }
    let bInt = b.reduce(0) { $0 * 2 + Int(String($1), radix: 2)! }
    let resultInt = aInt ^ bInt
    let resultString = String(resultInt, radix: 2)
    return resultString
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aInt = a.reduce(0) { $0 * 2 + Int(String($1), radix: 2)! }
    let bInt = b.reduce(0) { $0 * 2 + Int(String($1), radix: 2)! }
    let resultInt = aInt ^ bInt
    let resultString = String(resultInt, radix: 2)
    return resultString
}
```

func stringXor(a: String, b: String) -> String {
    let aChars = Array(a.characters)
    let bChars = Array(b.characters)
    
    guard aChars.count == bChars.count else {
        return ""
    }
    
    var resultChars = [Character]()
    
    for i in 0..<aChars.count {
        let aChar = aChars[i]
        let bChar = bChars[i]
        
        let xorResult = Character(aChar.unicodeScalars.first! ^ bChar.unicodeScalars.first!)
        resultChars.append(xorResult)
    }
    
    return String(resultChars)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aChars = Array(a.characters)
    let bChars = Array(b.characters)
    
    guard aChars.count == bChars.count else {
        return ""
    }
    
    var resultChars = [Character]()
    
    for i in 0..<aChars.count {
        let aChar = aChars[i]
        let bChar = bChars[i]
        
        let xorResult = Character(aChar.unicodeScalars.first! ^ bChar.unicodeScalars.first!)
        resultChars.append(xorResult)
    }
    
    return String(resultChars)
}
```

func stringXor(a: String, b: String) -> String {
    let aAsBits = a.map { String(format: "%04b", $0) }
    let bAsBits = b.map { String(format: "%04b", $0) }
    let result = aAsBits.enumerated().map { (index, bit) in
        let aBit = bit
        let bBit = bAsBits[index]
        let resultBit = aBit == "1"? (bBit == "1"? "0" : "1") : (bBit == "1"? "1" : "0")
        return String(format: "%04b", resultBit)
    }
    return result.joined()
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aAsBits = a.map { String(format: "%04b", $0) }
    let bAsBits = b.map { String(format: "%04b", $0) }
    let result = aAsBits.enumerated().map { (index, bit) in
        let aBit = bit
        let bBit = bAsBits[index]
        let resultBit = aBit == "1"? (bBit == "1"? "0" : "1") : (bBit == "1"? "1" : "0")
        return String(format: "%04b", resultBit)
    }
    return result.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aAsBits = a.map { String($0) }
    let bAsBits = b.map { String($0) }
    
    let result = aAsBits.enumerated().map { index, bit in
        let aBit = bit == "1"? 1 : 0
        let bBit = bAsBits[index] == "1"? 1 : 0
        return (aBit ^ bBit) == 1? "1" : "0"
    }
    
    return result.joined()
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aAsBits = a.map { String($0) }
    let bAsBits = b.map { String($0) }
    
    let result = aAsBits.enumerated().map { index, bit in
        let aBit = bit == "1"? 1 : 0
        let bBit = bAsBits[index] == "1"? 1 : 0
        return (aBit ^ bBit) == 1? "1" : "0"
    }
    
    return result.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aChars = a.characters.map { String(format: "%04b", $0.value) }
    let bChars = b.characters.map { String(format: "%04b", $0.value) }
    
    var resultChars = [Character]()
    for i in 0..<aChars.count {
        let aChar = aChars[i]
        let bChar = bChars[i]
        
        let aInt = Int(aChar, radix: 2)!
        let bInt = Int(bChar, radix: 2)!
        
        let resultInt = aInt ^ bInt
        let resultChar = String(format: "%04b", resultInt)
        
        resultChars.append(Character(resultChar))
    }
    
    return String(resultChars)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aChars = a.characters.map { String(format: "%04b", $0.value) }
    let bChars = b.characters.map { String(format: "%04b", $0.value) }
    
    var resultChars = [Character]()
    for i in 0..<aChars.count {
        let aChar = aChars[i]
        let bChar = bChars[i]
        
        let aInt = Int(aChar, radix: 2)!
        let bInt = Int(bChar, radix: 2)!
        
        let resultInt = aInt ^ bInt
        let resultChar = String(format: "%04b", resultInt)
        
        resultChars.append(Character(resultChar))
    }
    
    return String(resultChars)
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = a.map { String(format: "%04b", $0) }
    let bBinary = b.map { String(format: "%04b", $0) }
    let resultBinary = aBinary.enumerated().map { index, value in
        let aBit = value.first?? "0"
        let bBit = bBinary[index].first?? "0"
        return String(format: "%04b", Int(aBit) ^ Int(bBit))
    }
    return resultBinary.joined()
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = a.map { String(format: "%04b", $0) }
    let bBinary = b.map { String(format: "%04b", $0) }
    let resultBinary = aBinary.enumerated().map { index, value in
        let aBit = value.first?? "0"
        let bBit = bBinary[index].first?? "0"
        return String(format: "%04b", Int(aBit) ^ Int(bBit))
    }
    return resultBinary.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = a.map { String(format: "%08b", $0) }
    let bBinary = b.map { String(format: "%08b", $0) }
    let resultBinary = aBinary.enumerated().map { index, value in
        let bit = value.bit(at: index)
        return bit == 1? "0" : "1"
    }
    return String(format: "%08b", resultBinary.reduce(0, +))
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = a.map { String(format: "%08b", $0) }
    let bBinary = b.map { String(format: "%08b", $0) }
    let resultBinary = aBinary.enumerated().map { index, value in
        let bit = value.bit(at: index)
        return bit == 1? "0" : "1"
    }
    return String(format: "%08b", resultBinary.reduce(0, +))
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aChar = aArray[i]
        let bChar = bArray[i]
        let xorChar = aChar == bChar? "0" : "1"
        result.append(Character(xorChar))
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aChar = aArray[i]
        let bChar = bArray[i]
        let xorChar = aChar == bChar? "0" : "1"
        result.append(Character(xorChar))
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aChar = aArray[i]
        let bChar = bArray[i]
        let xorChar = aChar == bChar? "0" : "1"
        result.append(Character(xorChar))
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aChar = aArray[i]
        let bChar = bArray[i]
        let xorChar = aChar == bChar? "0" : "1"
        result.append(Character(xorChar))
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()

    for i in 0..<aArray.count {
        let aChar = aArray[i]
        let bChar = bArray[i]
        let xorChar = aChar == bChar? "0" : "1"
        result.append(Character(xorChar))
    }

    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()

    for i in 0..<aArray.count {
        let aChar = aArray[i]
        let bChar = bArray[i]
        let xorChar = aChar == bChar? "0" : "1"
        result.append(Character(xorChar))
    }

    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()

    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))?? 0
        let bInt = Int(String(bArray[i]))?? 0
        let xor = aInt ^ bInt
        result.append(Character(UnicodeScalar(xor)!))
    }

    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()

    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))?? 0
        let bInt = Int(String(bArray[i]))?? 0
        let xor = aInt ^ bInt
        result.append(Character(UnicodeScalar(xor)!))
    }

    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = Array(a).map { String($0) }
    let bBinary = Array(b).map { String($0) }
    var result = ""
    for i in 0..<aBinary.count {
        let a = Int(aBinary[i])?? 0
        let b = Int(bBinary[i])?? 0
        let xor = a ^ b
        result.append(String(xor))
    }
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = Array(a).map { String($0) }
    let bBinary = Array(b).map { String($0) }
    var result = ""
    for i in 0..<aBinary.count {
        let a = Int(aBinary[i])?? 0
        let b = Int(bBinary[i])?? 0
        let xor = a ^ b
        result.append(String(xor))
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorResult = aInt ^ bInt
        result.append(Character(UnicodeScalar(xorResult)!))
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorResult = aInt ^ bInt
        result.append(Character(UnicodeScalar(xorResult)!))
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xor = aInt ^ bInt
        result.append(Character(UnicodeScalar(xor)!))
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xor = aInt ^ bInt
        result.append(Character(UnicodeScalar(xor)!))
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()

    for i in 0..<aArray.count {
        let aChar = aArray[i]
        let bChar = bArray[i]
        let xorChar = aChar == bChar? "0" : "1"
        result.append(Character(xorChar))
    }

    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()

    for i in 0..<aArray.count {
        let aChar = aArray[i]
        let bChar = bArray[i]
        let xorChar = aChar == bChar? "0" : "1"
        result.append(Character(xorChar))
    }

    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xor = aInt ^ bInt
        result.append(Character(UnicodeScalar(xor)!))
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xor = aInt ^ bInt
        result.append(Character(UnicodeScalar(xor)!))
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()

    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xor = aInt ^ bInt
        result.append(Character(UnicodeScalar(xor)!))
    }

    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()

    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xor = aInt ^ bInt
        result.append(Character(UnicodeScalar(xor)!))
    }

    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    
    let xorData = aData.xor(bData)
    
    return String(data: xorData, encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    
    let xorData = aData.xor(bData)
    
    return String(data: xorData, encoding:.utf8)!
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let aBytes = aData.map { $0.bytes.first! }
    let bBytes = bData.map { $0.bytes.first! }
    let resultBytes = aBytes.enumerated().map { index, byte in
        let a = byte.element; let b = bBytes[index].element; return a ^ b
    }
    let resultData = Data(bytes: resultBytes)
    let resultString = String(data: resultData, encoding:.utf8)
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let aBytes = aData.map { $0.bytes.first! }
    let bBytes = bData.map { $0.bytes.first! }
    let resultBytes = aBytes.enumerated().map { index, byte in
        let a = byte.element; let b = bBytes[index].element; return a ^ b
    }
    let resultData = Data(bytes: resultBytes)
    let resultString = String(data: resultData, encoding:.utf8)
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let resultData = aData.xor(bData)
    return String(data: resultData, encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let resultData = aData.xor(bData)
    return String(data: resultData, encoding:.utf8)!
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    
    let aBytes = Array(aData.bytes)
    let bBytes = Array(bData.bytes)
    
    var resultBytes = [UInt8]()
    
    for i in 0..<aBytes.count {
        let aByte = aBytes[i]
        let bByte = bBytes[i]
        
        let xorByte = aByte ^ bByte
        resultBytes.append(xorByte)
    }
    
    let resultData = Data(bytes: resultBytes)
    let resultString = String(data: resultData, encoding:.utf8Bit)?? ""
    
    return resultString
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    
    let aBytes = Array(aData.bytes)
    let bBytes = Array(bData.bytes)
    
    var resultBytes = [UInt8]()
    
    for i in 0..<aBytes.count {
        let aByte = aBytes[i]
        let bByte = bBytes[i]
        
        let xorByte = aByte ^ bByte
        resultBytes.append(xorByte)
    }
    
    let resultData = Data(bytes: resultBytes)
    let resultString = String(data: resultData, encoding:.utf8Bit)?? ""
    
    return resultString
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    
    let aBytes = aData.map { $0.bytes }
    let bBytes = bData.map { $0.bytes }
    
    let maxBytes = max(aBytes.count, bBytes.count)
    
    var aBytes padded = aBytes
    var bBytes padded = bBytes
    
    if aBytes.count < maxBytes {
        aBytes.append(repeating: 0, count: maxBytes - aBytes.count)
    } else {
        bBytes.append(repeating: 0, count: maxBytes - bBytes.count)
    }
    
    let resultBytes = zip(aBytes, bBytes).map { $0 ^ $1 }
    
    let resultData = Data(bytes: resultBytes)
    let result = String(data: resultData, encoding:.utf8Bit)
    
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    
    let aBytes = aData.map { $0.bytes }
    let bBytes = bData.map { $0.bytes }
    
    let maxBytes = max(aBytes.count, bBytes.count)
    
    var aBytes padded = aBytes
    var bBytes padded = bBytes
    
    if aBytes.count < maxBytes {
        aBytes.append(repeating: 0, count: maxBytes - aBytes.count)
    } else {
        bBytes.append(repeating: 0, count: maxBytes - bBytes.count)
    }
    
    let resultBytes = zip(aBytes, bBytes).map { $0 ^ $1 }
    
    let resultData = Data(bytes: resultBytes)
    let result = String(data: resultData, encoding:.utf8Bit)
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let aBytes = aData.map { $0.bytes.first! }
    let bBytes = bData.map { $0.bytes.first! }
    let resultBytes = aBytes.enumerated().map { index, byte in
        let a = byte.element, let b = bBytes[index].element, let c = a ^ b {
            return c
        }
    }
    let resultData = Data(bytes: resultBytes)
    let result = String(data: resultData, encoding:.utf8Bit)
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let aBytes = aData.map { $0.bytes.first! }
    let bBytes = bData.map { $0.bytes.first! }
    let resultBytes = aBytes.enumerated().map { index, byte in
        let a = byte.element, let b = bBytes[index].element, let c = a ^ b {
            return c
        }
    }
    let resultData = Data(bytes: resultBytes)
    let result = String(data: resultData, encoding:.utf8Bit)
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let aBytes = aData.map { $0.bytes }
    let bBytes = bData.map { $0.bytes }
    let resultBytes = aBytes.enumerated().map { index, byte in
        let a = byte.element
        let b = bBytes[index].element
        return a ^ b
    }
    let resultData = Data(bytes: resultBytes)
    let result = String(data: resultData, encoding:.utf8Bit)
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let aBytes = aData.map { $0.bytes }
    let bBytes = bData.map { $0.bytes }
    let resultBytes = aBytes.enumerated().map { index, byte in
        let a = byte.element
        let b = bBytes[index].element
        return a ^ b
    }
    let resultData = Data(bytes: resultBytes)
    let result = String(data: resultData, encoding:.utf8Bit)
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    
    let aBytes = aData.map { $0.bytes }
    let bBytes = bData.map { $0.bytes }
    
    let resultBytes = aBytes.enumerated().map { index, byte in
        let a = byte.element, let b = bBytes[index].element in
        return a ^ b
    }
    
    let resultData = Data(bytes: resultBytes)
    let resultString = String(data: resultData, encoding:.utf8Bit)
    
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    
    let aBytes = aData.map { $0.bytes }
    let bBytes = bData.map { $0.bytes }
    
    let resultBytes = aBytes.enumerated().map { index, byte in
        let a = byte.element, let b = bBytes[index].element in
        return a ^ b
    }
    
    let resultData = Data(bytes: resultBytes)
    let resultString = String(data: resultData, encoding:.utf8Bit)
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    
    let xorData = aData.xor(bData)
    
    return String(data: xorData, encoding:.utf8)!
}

extension Data {
    func xor(_ other: Data) -> Data {
        var result = [UInt8](repeating: 0, count: self.count)
        
        for i in 0..<self.count {
            result[i] = self[i] ^ other[i]
        }
        
        return Data(bytes: result)
    }
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    
    let xorData = aData.xor(bData)
    
    return String(data: xorData, encoding:.utf8)!
}

extension Data {
    func xor(_ other: Data) -> Data {
        var result = [UInt8](repeating: 0, count: self.count)
        
        for i in 0..<self.count {
            result[i] = self[i] ^ other[i]
        }
        
        return Data(bytes: result)
    }
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let aBytes = Array(aData)
    let bBytes = Array(bData)
    let maxBytes = max(aBytes.count, bBytes.count)
    let aPadded = aBytes + Array(repeating: 0, count: maxBytes - aBytes.count)
    let bPadded = bBytes + Array(repeating: 0, count: maxBytes - bBytes.count)
    let resultBytes = zip(aPadded, bPadded).map { $0 ^ $1 }
    let resultData = Data(bytes: resultBytes)
    let resultString = String(data: resultData, encoding:.utf8Bit)
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let aBytes = Array(aData)
    let bBytes = Array(bData)
    let maxBytes = max(aBytes.count, bBytes.count)
    let aPadded = aBytes + Array(repeating: 0, count: maxBytes - aBytes.count)
    let bPadded = bBytes + Array(repeating: 0, count: maxBytes - bBytes.count)
    let resultBytes = zip(aPadded, bPadded).map { $0 ^ $1 }
    let resultData = Data(bytes: resultBytes)
    let resultString = String(data: resultData, encoding:.utf8Bit)
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.zip(bData).map { $0 ^ $1 }
    return String(data: Data(xorResult), encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.zip(bData).map { $0 ^ $1 }
    return String(data: Data(xorResult), encoding:.utf8)!
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.zip(bData).map { $0 ^ $1 }
    return String(data: Data(bytes: xorResult), encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.zip(bData).map { $0 ^ $1 }
    return String(data: Data(bytes: xorResult), encoding:.utf8)!
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let resultData = aData.xor(bData)
    return String(data: resultData, encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let resultData = aData.xor(bData)
    return String(data: resultData, encoding:.utf8)!
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.xor(bData)
    return String(data: xorResult, encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.xor(bData)
    return String(data: xorResult, encoding:.utf8)!
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.zip(bData).map { $0 ^ $1 }
    return String(data: Data(bytes: xorResult), encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.zip(bData).map { $0 ^ $1 }
    return String(data: Data(bytes: xorResult), encoding:.utf8)!
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let resultData = aData.xor(bData)
    return String(data: resultData, encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let resultData = aData.xor(bData)
    return String(data: resultData, encoding:.utf8)!
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.zip(bData).map { $0 ^ $1 }
    return String(data: Data(bytes: xorResult), encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.zip(bData).map { $0 ^ $1 }
    return String(data: Data(bytes: xorResult), encoding:.utf8)!
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let resultData = aData.xor(bData)
    return String(data: resultData, encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let resultData = aData.xor(bData)
    return String(data: resultData, encoding:.utf8)!
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.zip(bData).map { $0 ^ $1 }
    return String(data: Data(bytes: xorResult), encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let xorResult = aData.zip(bData).map { $0 ^ $1 }
    return String(data: Data(bytes: xorResult), encoding:.utf8)!
}
```

func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let resultData = aData.xor(bData)
    return String(data: resultData, encoding:.utf8)!
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aData = Data(a.utf8)
    let bData = Data(b.utf8)
    let resultData = aData.xor(bData)
    return String(data: resultData, encoding:.utf8)!
}
```

import Foundation

func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

```swift
import Foundation

/* 
 Function to multiply two matrices represented as 2D arrays of Doubles.
 Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
 */
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

/*
 Input are two strings a and b consisting only of 1s and 0s.
 Perform binary XOR on these inputs and return result also as a string.
 stringXor(a: "010", b: "110") == "100"
 */
func stringXor(a: String, b: String) -> String {
    
    
    var result = [[Double]()
    for

import Foundation

func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

```swift
import Foundation

/* 
 Function to multiply two matrices represented as 2D arrays of Doubles.
 Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
 */
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

/*
 Input are two strings a and b consisting only of 1s and 0s.
 Perform binary XOR on these inputs and return result also as a string.
 stringXor(a: "010", b: "110") == "100"
 */
func stringXor(a: String, b: String) -> String {

import Foundation

func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

```swift
import Foundation

/* 
 Function to multiply two matrices represented as 2D arrays of Doubles.
 Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
 */
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

/*
 Input are two strings a and b consisting only of 1s and 0s.
 Perform binary XOR on these inputs and return result also as a string.
 stringXor(a: "010", b: "110") == "100"
 */
func stringXor(a: String, b: String) -> String {
    
    
    
    

import Foundation

func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

```swift
import Foundation

/* 
 Function to multiply two matrices represented as 2D arrays of Doubles.
 Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
 */
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

/*
 Input are two strings a and b consisting only of 1s and 0s.
 Perform binary XOR on these inputs and return result also as a string.
 stringXor(a: "010", b: "110") == "100"
 */
func stringXor(a: String, b: String) -> String {
    
    
    
    
    
    var result = [[

import Foundation

func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

```swift
import Foundation

/* 
 Function to multiply two matrices represented as 2D arrays of Doubles.
 Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
 */
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

/*
 Input are two strings a and b consisting only of 1s and 0s.
 Perform binary XOR on these inputs and return result also as a string.
 stringXor(a: "010", b: "110") == "100"
 */
func stringXor(a: String, b: String) -> String {
    
    
    

import Foundation

func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

```swift
import Foundation

/* 
 Function to multiply two matrices represented as 2D arrays of Doubles.
 Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
 */
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

/*
 Input are two strings a and b consisting only of 1s and 0s.
 Perform binary XOR on these inputs and return result also as a string.
 stringXor(a: "010", b: "110") == "100"
 */
func stringXor(a: String, b: String) -> String {
    
    

import Foundation

func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

```swift
import Foundation

/* 
 Function to multiply two matrices represented as 2D arrays of Doubles.
 Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
 */
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

/*
 Input are two strings a and b consisting only of 1s and 0s.
 Perform binary XOR on these inputs and return result also as a string.
 stringXor(a: "010", b: "110") == "100"
 */
func stringXor(a: String, b: String) -> String {
    
    
    
    var result = [[Double]()
    for (0
    for (0
    

import Foundation

func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

```swift
import Foundation

/* 
 Function to multiply two matrices represented as 2D arrays of Doubles.
 Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
 */
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

/*
 Input are two strings a and b consisting only of 1s and 0s.
 Perform binary XOR on these inputs and return result also as a string.
 stringXor(a: "010", b: "110") == "100"
 */
func stringXor(a: String, b: String) -> String {

import Foundation

func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

```swift
import Foundation

/* 
 Function to multiply two matrices represented as 2D arrays of Doubles.
 Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
 */
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

/*
 Input are two strings a and b consisting only of 1s and 0s.
 Perform binary XOR on these inputs and return result also as a string.
 stringXor(a: "010", b: "110") == "100"
 */
func stringXor(a: String, b: String) -> String {

import Foundation

func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

```swift
import Foundation

/* 
 Function to multiply two matrices represented as 2D arrays of Doubles.
 Returns a new matrix that is the product of `lhs` and `rhs`, or `nil` if the matrices cannot be multiplied.
 */
func multiply(_ lhs: [[Double]], _ rhs: [[Double]]) -> [[Double]]? {
    guard lhs[0].count == rhs.count else { 
        return nil 
    }
    
    var result = [[Double]](repeating: [Double](repeating: 0, count: rhs[0].count), count: lhs.count)
    for i in 0..<lhs.count {
        for j in 0..<rhs[0].count {
            for k in 0..<rhs.count {
                result[i][j] += lhs[i][k] * rhs[k][j]
            }
        }
    }
    
    return result
}

/*
 Input are two strings a and b consisting only of 1s and 0s.
 Perform binary XOR on these inputs and return result also as a string.
 stringXor(a: "010", b: "110") == "100"
 */
func stringXor(a: String, b: String) -> String {
    
    
    
    var result = [[Double](repeekse
    for i in
    for (a

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    var a.append(a.unicodeScalars = a.unicodeScalars.unicodeScalars.filter {
    

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    
    var a = a.map {

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    
    var a = a.map {
        // Convert strings are the inputs to character arrays of UInt, convert them to integer arrays
        let a = a = String and b to integer arrays, perform xor ""
       .unicodeScalars,
        let a = a = a = a = a.utf3276340
        Int32761
        b

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros on the left
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    } else if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros on the left
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    } else if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

print(stringXor(a: "010", b: "110")) // Output: "100"
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a.count > i ? String(a[a.index(a.startIndex, offsetBy: i)]) : ""
        let bBit = b.count > i ? String(b[b.index(b.startIndex, offsetBy: i)]) : ""
        
        let xorResult = aBit == bBit ? "0" : "1"
        result.append(xorResult)
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a.count > i ? String(a[a.index(a.startIndex, offsetBy: i)]) : ""
        let bBit = b.count > i ? String(b[b.index(b.startIndex, offsetBy: i)]) : ""
        
        let xorResult = aBit == bBit ? "0" : "1"
        result.append(xorResult)
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    var aIndex = a.count - 1
    var bIndex = b.count - 1
    
    var carry = 0
    
    while aIndex >= 0 || bIndex >= 0 || carry > 0 {
        let aDigit = aIndex >= 0 ? Int(String(a[aIndex])) ?? 0 : 0
        let bDigit = bIndex >= 0 ? Int(String(b[bIndex])) ?? 0 : 0
        
        let xorResult = aDigit != bDigit
        let carryOut = carry + xorResult
        
        result.insert(String(carryOut % 2), at: 0)
        
        carry = carryOut / 2
        
        aIndex -= 1
        bIndex -= 1
    }
    
    return result.reversed()
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    var aIndex = a.count - 1
    var bIndex = b.count - 1
    
    var carry = 0
    
    while aIndex >= 0 || bIndex >= 0 || carry > 0 {
        let aDigit = aIndex >= 0 ? Int(String(a[aIndex])) ?? 0 : 0
        let bDigit = bIndex >= 0 ? Int(String(b[bIndex])) ?? 0 : 0
        
        let xorResult = aDigit != bDigit
        let carryOut = carry + xorResult
        
        result.insert(String(carryOut % 2), at: 0)
        
        carry = carryOut / 2
        
        aIndex -= 1
        bIndex -= 1
    }
    
    return result.reversed()
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a.index(a.startIndex, offsetBy: i) == "1" ? "1" : "0"
        let bBit = b.index(b.startIndex, offsetBy: i) == "1" ? "1" : "0"
        result.append(aBit == bBit ? "0" : "1")
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a.index(a.startIndex, offsetBy: i) == "1" ? "1" : "0"
        let bBit = b.index(b.startIndex, offsetBy: i) == "1" ? "1" : "0"
        result.append(aBit == bBit ? "0" : "1")
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    }
    if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    }
    if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a.count > i ? String(a[a.index(a.startIndex, offsetBy: i)]) : ""
        let bBit = b.count > i ? String(b[b.index(b.startIndex, offsetBy: i)]) : ""
        
        let xorResult = aBit == bBit ? "0" : "1"
        result.append(xorResult)
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a.count > i ? String(a[a.index(a.startIndex, offsetBy: i)]) : ""
        let bBit = b.count > i ? String(b[b.index(b.startIndex, offsetBy: i)]) : ""
        
        let xorResult = aBit == bBit ? "0" : "1"
        result.append(xorResult)
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a.count > i ? String(a[a.index(a.startIndex, offsetBy: i)]) : ""
        let bBit = b.count > i ? String(b[b.index(b.startIndex, offsetBy: i)]) : ""
        
        let xorResult = aBit == bBit ? "0" : "1"
        result.append(xorResult)
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a.count > i ? String(a[a.index(a.startIndex, offsetBy: i)]) : ""
        let bBit = b.count > i ? String(b[b.index(b.startIndex, offsetBy: i)]) : ""
        
        let xorResult = aBit == bBit ? "0" : "1"
        result.append(xorResult)
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    }
    if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        let aBit = a[String(a[i])]
        let bBit = b[String(b[i])]
        if aBit == bBit {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    }
    if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        let aBit = a[String(a[i])]
        let bBit = b[String(b[i])]
        if aBit == bBit {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    var aIndex = a.count - 1
    var bIndex = b.count - 1
    
    var carry = 0
    repeat {
        let aBit = Int(String(a[aIndex]))
        let bBit = Int(String(b[bIndex]))
        
        let xorResult = aBit != bBit ? (aBit + bBit + carry) % 2 : carry
        carry = (aBit + bBit + carry) >= 2 ? 1 : 0
        
        result.append(String(xorResult))
        
        aIndex -= 1
        bIndex -= 1
    } while aIndex >= 0 || bIndex >= 0 || carry != 0
    
    return result.reversed()
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    var aIndex = a.count - 1
    var bIndex = b.count - 1
    
    var carry = 0
    repeat {
        let aBit = Int(String(a[aIndex]))
        let bBit = Int(String(b[bIndex]))
        
        let xorResult = aBit != bBit ? (aBit + bBit + carry) % 2 : carry
        carry = (aBit + bBit + carry) >= 2 ? 1 : 0
        
        result.append(String(xorResult))
        
        aIndex -= 1
        bIndex -= 1
    } while aIndex >= 0 || bIndex >= 0 || carry != 0
    
    return result.reversed()
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    var aIndex = a.count - 1
    var bIndex = b.count - 1
    
    var carry = 0
    repeat {
        let aBit = Int(String(a[aIndex]))
        let bBit = Int(String(b[bIndex]))
        
        let xorResult = aBit != bBit ? (aBit + bBit + carry) % 2 : carry
        carry = (aBit + bBit + carry) >= 2 ? 1 : 0
        
        result.append(String(xorResult))
        
        aIndex -= 1
        bIndex -= 1
    } while aIndex >= 0 || bIndex >= 0 || carry != 0
    
    return result.reversed()
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    var aIndex = a.count - 1
    var bIndex = b.count - 1
    
    var carry = 0
    repeat {
        let aBit = Int(String(a[aIndex]))
        let bBit = Int(String(b[bIndex]))
        
        let xorResult = aBit != bBit ? (aBit + bBit + carry) % 2 : carry
        carry = (aBit + bBit + carry) >= 2 ? 1 : 0
        
        result.append(String(xorResult))
        
        aIndex -= 1
        bIndex -= 1
    } while aIndex >= 0 || bIndex >= 0 || carry != 0
    
    return result.reversed()
}

print(stringXor(a: "010", b: "110")) // Output: "100"
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    let aArray = Array(a.pad(toLength: maxLength, withHead: "0"))
    let bArray = Array(b.pad(toLength: maxLength, withHead: "0"))
    for i in 0..<maxLength {
        let aBit = aArray[aArray.count - i - 1] == "1"
        let bBit = bArray[bArray.count - i - 1] == "1"
        if aBit ^ bBit {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    let aArray = Array(a.pad(toLength: maxLength, withHead: "0"))
    let bArray = Array(b.pad(toLength: maxLength, withHead: "0"))
    for i in 0..<maxLength {
        let aBit = aArray[aArray.count - i - 1] == "1"
        let bBit = bArray[bArray.count - i - 1] == "1"
        if aBit ^ bBit {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.padLeading(to: maxLength, with: "0")
    var b = b.padLeading(to: maxLength, with: "0")
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a[a.index(a.startIndex, offsetBy: i)] == "1"
        let bBit = b[b.index(b.startIndex, offsetBy: i)] == "1"
        result += aBit ^ bBit ? "1" : "0"
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.padLeading(to: maxLength, with: "0")
    var b = b.padLeading(to: maxLength, with: "0")
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a[a.index(a.startIndex, offsetBy: i)] == "1"
        let bBit = b[b.index(b.startIndex, offsetBy: i)] == "1"
        result += aBit ^ bBit ? "1" : "0"
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.repeating("0").padded(toLength: maxLength, withPad: "0")
    var b = b.repeating("0").padded(toLength: maxLength, withPad: "0")
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = a[a.index(a.startIndex, offsetBy: i)] == "1"
        let bBit = b[b.index(b.startIndex, offsetBy: i)] == "1"
        result += aBit ^ bBit ? "1" : "0"
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.repeating("0").padded(toLength: maxLength, withPad: "0")
    var b = b.repeating("0").padded(toLength: maxLength, withPad: "0")
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = a[a.index(a.startIndex, offsetBy: i)] == "1"
        let bBit = b[b.index(b.startIndex, offsetBy: i)] == "1"
        result += aBit ^ bBit ? "1" : "0"
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.padLeading(to: maxLength, with: "0")
    var b = b.padLeading(to: maxLength, with: "0")
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = a[a.index(a.startIndex, offsetBy: i)] == "1"
        let bBit = b[b.index(b.startIndex, offsetBy: i)] == "1"
        result += aBit ^ bBit ? "1" : "0"
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.padLeading(to: maxLength, with: "0")
    var b = b.padLeading(to: maxLength, with: "0")
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = a[a.index(a.startIndex, offsetBy: i)] == "1"
        let bBit = b[b.index(b.startIndex, offsetBy: i)] == "1"
        result += aBit ^ bBit ? "1" : "0"
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.padLeading(to: maxLength, with: "0")
    var b = b.padLeading(to: maxLength, with: "0")
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = a[a.index(a.startIndex, offsetBy: i)] == "1"
        let bBit = b[b.index(b.startIndex, offsetBy: i)] == "1"
        result += aBit ^ bBit ? "1" : "0"
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.padLeading(to: maxLength, with: "0")
    var b = b.padLeading(to: maxLength, with: "0")
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = a[a.index(a.startIndex, offsetBy: i)] == "1"
        let bBit = b[b.index(b.startIndex, offsetBy: i)] == "1"
        result += aBit ^ bBit ? "1" : "0"
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aArray = Array(a)
    let bArray = Array(b)
    let maxLength = max(aArray.count, bArray.count)
    
    for i in 0..<maxLength {
        let aIndex = aArray.count - 1 - i
        let bIndex = bArray.count - 1 - i
        let aBit = aIndex >= 0 ? Int(aArray[aIndex].description) : 0
        let bBit = bIndex >= 0 ? Int(bArray[bIndex].description) : 0
        result = String(aBit ^ bBit) + result
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aArray = Array(a)
    let bArray = Array(b)
    let maxLength = max(aArray.count, bArray.count)
    
    for i in 0..<maxLength {
        let aIndex = aArray.count - 1 - i
        let bIndex = bArray.count - 1 - i
        let aBit = aIndex >= 0 ? Int(aArray[aIndex].description) : 0
        let bBit = bIndex >= 0 ? Int(bArray[bIndex].description) : 0
        result = String(aBit ^ bBit) + result
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.repeating("0").padding(toLength: maxLength, withPad: "0")
    var b = b.repeating("0").padding(toLength: maxLength, withPad: "0")
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = Int(String(Array(a)[i]))
        let bBit = Int(String(Array(b)[i]))
        
        if (aBit! ^ bBit!) == 1 {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.repeating("0").padding(toLength: maxLength, withPad: "0")
    var b = b.repeating("0").padding(toLength: maxLength, withPad: "0")
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = Int(String(Array(a)[i]))
        let bBit = Int(String(Array(b)[i]))
        
        if (aBit! ^ bBit!) == 1 {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    let aArray = Array(a.reversed())
    let bArray = Array(b.reversed())
    
    for i in 0..<maxLength {
        let aBit = i < a.count ? aArray[i] == "1" : false
        let bBit = i < b.count ? bArray[i] == "1" : false
        
        if aBit != bBit {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    
    return String(result.reversed())
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    let aArray = Array(a.reversed())
    let bArray = Array(b.reversed())
    
    for i in 0..<maxLength {
        let aBit = i < a.count ? aArray[i] == "1" : false
        let bBit = i < b.count ? bArray[i] == "1" : false
        
        if aBit != bBit {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    
    return String(result.reversed())
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.repeating("0").padding(toLength: maxLength, withPad: "0")
    var b = b.repeating("0").padding(toLength: maxLength, withPad: "0")
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = Int(String(Array(a)[i]))
        let bBit = Int(String(Array(b)[i]))
        
        if (aBit! ^ bBit!) == 1 {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var a = a.repeating("0").padding(toLength: maxLength, withPad: "0")
    var b = b.repeating("0").padding(toLength: maxLength, withPad: "0")
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = Int(String(Array(a)[i]))
        let bBit = Int(String(Array(b)[i]))
        
        if (aBit! ^ bBit!) == 1 {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aArray = Array(a)
    let bArray = Array(b)
    
    for i in 0..<max(aArray.count, bArray.count) {
        let aBit = aArray.count > i ? Int(aArray[aArray.count - i - 1] - '0') : 0
        let bBit = bArray.count > i ? Int(bArray[bArray.count - i - 1] - '0') : 0
        result = String(aBit ^ bBit) + result
    }
    
    return result
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aArray = Array(a)
    let bArray = Array(b)
    
    for i in 0..<max(aArray.count, bArray.count) {
        let aBit = aArray.count > i ? Int(aArray[aArray.count - i - 1] - '0') : 0
        let bBit = bArray.count > i ? Int(bArray[bArray.count - i - 1] - '0') : 0
        result = String(aBit ^ bBit) + result
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    let xorResult = aBinary.map(bBinary, { $0 ^ $1 })
    return String(xorResult, radix: 2)
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    let xorResult = aBinary.map(bBinary, { $0 ^ $1 })
    return String(xorResult, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    let xorResult = aBinary.map { $0 ^ bBinary.map { $0 } }
    return String(xorResult, radix: 2)
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    let xorResult = aBinary.map { $0 ^ bBinary.map { $0 } }
    return String(xorResult, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    let xorResult = aBinary.map(bBinary, { $0 ^ $1 })
    return String(xorResult, radix: 2)
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    let xorResult = aBinary.map(bBinary, { $0 ^ $1 })
    return String(xorResult, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    let xorResult = aBinary.map { $0 ^ bBinary.map { $0 } }
    return String(xorResult, radix: 2)
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    let xorResult = aBinary.map { $0 ^ bBinary.map { $0 } }
    return String(xorResult, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    
    let xorResult = aBinary.map { $0 ^ bBinary.map { $0 } }
    
    return String(xorResult, radix: 2)
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    
    let xorResult = aBinary.map { $0 ^ bBinary.map { $0 } }
    
    return String(xorResult, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    let xorResult = aBinary.map { $0 ^ bBinary }
    return String(xorResult, radix: 2)
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    let xorResult = aBinary.map { $0 ^ bBinary }
    return String(xorResult, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    
    let xorResult = aBinary.map { $0 ^ bBinary }
    
    return String(xorResult, radix: 2)
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    
    let xorResult = aBinary.map { $0 ^ bBinary }
    
    return String(xorResult, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    
    let xorResult = aBinary.map { $0 ^ bBinary.map { $0 } }
    
    return String(xorResult, radix: 2)
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = UInt(a, radix: 2)
    let bBinary = UInt(b, radix: 2)
    
    let xorResult = aBinary.map { $0 ^ bBinary.map { $0 } }
    
    return String(xorResult, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    let binaryA = Array(a.reversed()).map { $0 == "1" ? 1 : 0 }
    let binaryB = Array(b.reversed()).map { $0 == "1" ? 1 : 0 }
    
    var result = [Int]()
    let maxLength = max(binaryA.count, binaryB.count)
    
    for i in 0..<maxLength {
        let bitA = i < binaryA.count ? binaryA[i] : 0
        let bitB = i < binaryB.count ? binaryB[i] : 0
        result.append(bitA ^ bitB)
    }
    
    return result.map { $0 == 1 ? "1" : "0" }.reversed().joined()
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let binaryA = Array(a.reversed()).map { $0 == "1" ? 1 : 0 }
    let binaryB = Array(b.reversed()).map { $0 == "1" ? 1 : 0 }
    
    var result = [Int]()
    let maxLength = max(binaryA.count, binaryB.count)
    
    for i in 0..<maxLength {
        let bitA = i < binaryA.count ? binaryA[i] : 0
        let bitB = i < binaryB.count ? binaryB[i] : 0
        result.append(bitA ^ bitB)
    }
    
    return result.map { $0 == 1 ? "1" : "0" }.reversed().joined()
}
```

func stringXor(a: String, b: String) -> String {
    let binaryA = Array(a.reversed()).map { $0 == "1" ? 1 : 0 }
    let binaryB = Array(b.reversed()).map { $0 == "1" ? 1 : 0 }
    var result = [Int]()
    
    var i = 0
    var carryOver = 0
    
    while i < max(binaryA.count, binaryB.count) {
        let aBit = i < binaryA.count ? binaryA[i] : 0
        let bBit = i < binaryB.count ? binaryB[i] : 0
        let sum = aBit + bBit + carryOver
        
        result.append(sum % 2)
        carryOver = sum / 2
        
        i += 1
    }
    
    if carryOver > 0 {
        result.append(carryOver)
    }
    
    return result.reversed().map { String($0) }.joined()
}

<｜begin▁of▁sentence｜>You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let binaryA = Array(a.reversed()).map { $0 == "1" ? 1 : 0 }
    let binaryB = Array(b.reversed()).map { $0 == "1" ? 1 : 0 }
    var result = [Int]()
    
    var i = 0
    var carryOver = 0
    
    while i < max(binaryA.count, binaryB.count) {
        let aBit = i < binaryA.count ? binaryA[i] : 0
        let bBit = i < binaryB.count ? binaryB[i] : 0
        let sum = aBit + bBit + carryOver
        
        result.append(sum % 2)
        carryOver = sum / 2
        
        i += 1
    }
    
    if carryOver > 0 {
        result.append(carryOver)
    }
    
    return result.reversed().map { String($0) }.joined()
}
```

import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
import Foundation

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (charA, charB) in zip(a, b) {
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = ""
    
    for i in 0..<aArray.count {
        if aArray[i] == bArray[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = ""
    
    for i in 0..<aArray.count {
        if aArray[i] == bArray[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = String(a[a.index(a.startIndex, offsetBy: i)])
        let bChar = String(b[b.index(b.startIndex, offsetBy: i)])
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = String(a[a.index(a.startIndex, offsetBy: i)])
        let bChar = String(b[b.index(b.startIndex, offsetBy: i)])
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aCharacters = Array(a)
    let bCharacters = Array(b)
    var result = ""
    
    for i in 0..<aCharacters.count {
        if aCharacters[i] == bCharacters[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    let aCharacters = Array(a)
    let bCharacters = Array(b)
    var result = ""
    
    for i in 0..<aCharacters.count {
        if aCharacters[i] == bCharacters[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = String(a[a.index(a.startIndex, offsetBy: i)])
        let bChar = String(b[b.index(b.startIndex, offsetBy: i)])
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = String(a[a.index(a.startIndex, offsetBy: i)])
        let bChar = String(b[b.index(b.startIndex, offsetBy: i)])
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = String(a[a.index(a.startIndex, offsetBy: i)])
        let bChar = String(b[b.index(b.startIndex, offsetBy: i)])
        let xor = aChar == bChar ? "0" : "1"
        result += xor
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = String(a[a.index(a.startIndex, offsetBy: i)])
        let bChar = String(b[b.index(b.startIndex, offsetBy: i)])
        let xor = aChar == bChar ? "0" : "1"
        result += xor
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.characters.count {
        let aChar = a[a.startIndex.advancedBy(i)]
        let bChar = b[b.startIndex.advancedBy(i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.characters.count {
        let aChar = a[a.startIndex.advancedBy(i)]
        let bChar = b[b.startIndex.advancedBy(i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    
    var result = [Character]()
    
    for i in 0..<a.count {
        let a = a[i]
        let b = b[i]
        
        if a == b {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return String(result)
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    
    var result = [Character]()
    
    for i in 0..<a.count {
        let a = a[i]
        let b = b[i]
        
        if a == b {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.characters.count {
        let aChar = a.characters.dropFirst(i).first!
        let bChar = b.characters.dropFirst(i).first!
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.characters.count {
        let aChar = a.characters.dropFirst(i).first!
        let bChar = b.characters.dropFirst(i).first!
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    
    var result = [Character]()
    
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return String(result)
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    
    var result = [Character]()
    
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.characters.count {
        let aChar = a[a.startIndex.advancedBy(i)]
        let bChar = b[b.startIndex.advancedBy(i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.characters.count {
        let aChar = a[a.startIndex.advancedBy(i)]
        let bChar = b[b.startIndex.advancedBy(i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    
    var result = ""
    
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    
    var result = ""
    
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    
    var result = ""
    
    for i in 0..<a.count {
        let a = Int(String(a[i]))!
        let b = Int(String(b[i]))!
        
        if a == b {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    
    var result = ""
    
    for i in 0..<a.count {
        let a = Int(String(a[i]))!
        let b = Int(String(b[i]))!
        
        if a == b {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    
    var result = [Character]()
    
    for (i, _) in a.enumerated() {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return String(result)
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    
    var result = [Character]()
    
    for (i, _) in a.enumerated() {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i]!= b[i] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i]!= b[i] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for index in 0..<a.count {
        if a[index]!= b[index] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for index in 0..<a.count {
        if a[index]!= b[index] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for index in 0..<a.count {
        if a[index] == b[index] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for index in 0..<a.count {
        if a[index] == b[index] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    // Your code here.
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Your code here.
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        let aBit = a.count > i? a[i] : "0"
        let bBit = b.count > i? b[i] : "0"
        result.append(aBit == bBit? "0" : "1")
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        let aBit = a.count > i? a[i] : "0"
        let bBit = b.count > i? b[i] : "0"
        result.append(aBit == bBit? "0" : "1")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i]!= b[i] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i]!= b[i] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var a = a.unicodeScalars
    var b = b.unicodeScalars
    var result = [String]()
    for i in 0..<max(a.count, b.count) {
        if i < a.count {
            a[i] ^= b[i]
        }
        if i < b.count {
            b[i] ^= a[i]
        }
        if a[i] == 0 {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return String(result)
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var a = a.unicodeScalars
    var b = b.unicodeScalars
    var result = [String]()
    for i in 0..<max(a.count, b.count) {
        if i < a.count {
            a[i] ^= b[i]
        }
        if i < b.count {
            b[i] ^= a[i]
        }
        if a[i] == 0 {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    return a.map {
        if a == "0" {
            return b[0]
        } else {
            return b[1]
        }
    }.joined()
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    return a.map {
        if a == "0" {
            return b[0]
        } else {
            return b[1]
        }
    }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.characters.contains("1")[i]? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.characters.contains("1")[i]? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.characters.contains("1")[i]? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.characters.contains("1")[i]? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.characters.contains("1")[i]? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.characters.contains("1")[i]? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.characters.contains("1")[i]? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.characters.contains("1")[i]? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.characters.contains("1")[i]? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.characters.contains("1")[i]? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.characters.contains("1")[i]? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.characters.contains("1")[i]? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.characters.contains("1")[i]? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.characters.contains("1")[i]? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.characters.contains("1")[i]? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.characters.contains("1")[i]? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}

<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<|im_end|>
<|im_start|>assistant
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let maxLength = max(a.count, b.count)
    
    for i in 0..<maxLength {
        let aBit: Int
        let bBit: Int
        
        if i < a.count {
            aBit = a.charAt(a.index(a.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            aBit = 0
        }
        
        if i < b.count {
            bBit = b.charAt(b.index(b.startIndex, offsetBy: i)).isEqual("1")? 1 : 0
        } else {
            bBit = 0
        }
        
        let xorResult = aBit ^ bBit
        result.append(String(xorResult))
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            if b[index] == "1" {
                result += "0"
            } else {
                result += "1"
            }
        } else {
            result += "0"
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            if b[index] == "1" {
                result += "0"
            } else {
                result += "1"
            }
        } else {
            result += "0"
        }
    }
    return result
}
```
### Challenge

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result.append(b[index] == "1" ? "0" : "1")
        } else {
            result.append(b[index] == "1" ? "1" : "0")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result.append(b[index] == "1" ? "0" : "1")
        } else {
            result.append(b[index] == "1" ? "1" : "0")
        }
    }
    return result
}
```
### Problem

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result += b[index] == "1" ? "0" : "1"
        } else {
            result += b[index] == "0" ? "1" : "0"
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result += b[index] == "1" ? "0" : "1"
        } else {
            result += b[index] == "0" ? "1" : "0"
        }
    }
    return result
}
```
### Challenge

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == "1" && b[i] == "1" {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == "1" && b[i] == "1" {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```
### Problem

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result += b[index]
        } else {
            result += a[index]
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result += b[index]
        } else {
            result += a[index]
        }
    }
    return result
}
```
### Challenge

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result += b[index]
        } else {
            result += a[index]
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result += b[index]
        } else {
            result += a[index]
        }
    }
    return result
}
```
### Problem

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```
### Problem

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result += b[index]
        } else {
            result += a[index]
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result += b[index]
        } else {
            result += a[index]
        }
    }
    return result
}
```
### Problem

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result.append(b[index] == "1" ? "0" : "1")
        } else {
            result.append(b[index] == "1" ? "1" : "0")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == "1" {
            result.append(b[index] == "1" ? "0" : "1")
        } else {
            result.append(b[index] == "1" ? "1" : "0")
        }
    }
    return result
}
```
### Challenge

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[a.index(a.startIndex, offsetBy: i)]
        let charB = b[b.index(b.startIndex, offsetBy: i)]
        if charA == "1" && charB == "1" {
            result += "0"
        } else if charA == "0" && charB == "0" {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[a.index(a.startIndex, offsetBy: i)]
        let charB = b[b.index(b.startIndex, offsetBy: i)]
        if charA == "1" && charB == "1" {
            result += "0"
        } else if charA == "0" && charB == "0" {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}
```
### Challenge

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, c) in a.enumerated() {
        let d = b[i]
        result.append(c == d ? "0" : "1")
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, c) in a.enumerated() {
        let d = b[i]
        result.append(c == d ? "0" : "1")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = a.map { Int(String($0))! }
    let b = b.map { Int(String($0))! }
    let result = zip(a, b).map { $0 ^ $1 }
    return result.map { String($0) }.joined()
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let a = a.map { Int(String($0))! }
    let b = b.map { Int(String($0))! }
    let result = zip(a, b).map { $0 ^ $1 }
    return result.map { String($0) }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, c) in a.enumerated() {
        let d = b[i]
        result.append(String(c == d ? "0" : "1"))
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, c) in a.enumerated() {
        let d = b[i]
        result.append(String(c == d ? "0" : "1"))
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        if aChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, element) in a.enumerated() {
        let element = element == "1" ? "0" : "1"
        result += element
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, element) in a.enumerated() {
        let element = element == "1" ? "0" : "1"
        result += element
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    return String(zip(a, b).map { $0 == $1 ? "0" : "1" })
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    return String(zip(a, b).map { $0 == $1 ? "0" : "1" })
}
```
### Link

func stringXor(a: String, b: String) -> String {
    let a = a.map { Int(String($0))! }
    let b = b.map { Int(String($0))! }
    let result = zip(a, b).map { $0 ^ $1 }
    return result.map { String($0) }.joined()
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let a = a.map { Int(String($0))! }
    let b = b.map { Int(String($0))! }
    let result = zip(a, b).map { $0 ^ $1 }
    return result.map { String($0) }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    return String(a.map { Int(String($0))! ^ Int(String(b[$0]))! })
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    return String(a.map { Int(String($0))! ^ Int(String(b[$0]))! })
}
```
### Link

func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    let result = zip(a, b).map { $0 != $1 ? "1" : "0" }
    return String(result)
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    let result = zip(a, b).map { $0 != $1 ? "1" : "0" }
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        result.append(i == j ? "0" : "1")
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        result.append(i == j ? "0" : "1")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        result.append(i == j ? "0" : "1")
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        result.append(i == j ? "0" : "1")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" }
    let b = b.map { $0 == "1" }
    let result = zip(a, b).map { $0 != $1 }
    return result.map { $0 ? "1" : "0" }.joined()
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" }
    let b = b.map { $0 == "1" }
    let result = zip(a, b).map { $0 != $1 }
    return result.map { $0 ? "1" : "0" }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        if i == j {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        if i == j {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        if i == j {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        if i == j {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    var result = ""
    for i in 0..<a.count {
        result += a[i] == b[i] ? "0" : "1"
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    var result = ""
    for i in 0..<a.count {
        result += a[i] == b[i] ? "0" : "1"
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    let result = zip(a, b).map { $0 == $1 ? "0" : "1" }
    return result.joined()
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    let result = zip(a, b).map { $0 == $1 ? "0" : "1" }
    return result.joined()
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        result += i == j ? "0" : "1"
    }
    return result
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        result += i == j ? "0" : "1"
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    let result = a.enumerated().map { (i, char) -> String in
        return char == b[i] ? "0" : "1"
    }
    return result.joined()
}

### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    let result = a.enumerated().map { (i, char) -> String in
        return char == b[i] ? "0" : "1"
    }
    return result.joined()
}
```

func stringXor(a: String, b: String) -> String {
    // Check if the lengths of A and B are EQUAL
    guard A.count == b.count else {
        fatalError("Input strings must be of equal length.")
    }
    
    // Perform XOR operation
    var result = ""
    for (aChar, bChar) in zip(A, b) {
        if AChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Check if the lengths of A and B are EQUAL
    guard A.count == b.count else {
        fatalError("Input strings must be of equal length.")
    }
    
    // Perform XOR operation
    var result = ""
    for (aChar, bChar) in zip(A, b) {
        if AChar == bChar {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

// Example usage:
let result = stringXor(A: "010", b: "110")
print(result)  // Output: "100"
```

func stringXor(a: String, b: String) -> String {
    // Ensure both strings are of equal length by padding the shorter one with zeros.
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation by comparing each character.
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Ensure both strings are of equal length by padding the shorter one with zeros.
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation by comparing each character.
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

// Example usage:
let result = stringXor(A: "010", b: "110")
print(result)  // Output: "100"
```

This function takes two binary strings A and B, compares each character, and returns a new string where each character is the result of XOR operation on the corresponding characters of A and B.

func stringXor(a: String, b: String) -> String {
    // Check if the lengths of A and B are EQUAL
    if A.count != b.count {
        fatalError("Input strings must be of equal length.")
    }
    
    // Initialize an empty string to store the result
    var result = ""
    
    // Iterate over the characters of A and B
    for i in 0..<A.count {
        // Get the character at the current index of A and B
        let charA = String(A[A.index(A.startIndex, offsetBy: i)])
        let charB = String(b[b.index(b.startIndex, offsetBy: i)])
        
        // Perform XOR operation and append the result to the result string
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Check if the lengths of A and B are EQUAL
    if A.count != b.count {
        fatalError("Input strings must be of equal length.")
    }
    
    // Initialize an empty string to store the result
    var result = ""
    
    // Iterate over the characters of A and B
    for i in 0..<A.count {
        // Get the character at the current index of A and B
        let charA = String(A[A.index(A.startIndex, offsetBy: i)])
        let charB = String(b[b.index(b.startIndex, offsetBy: i)])
        
        // Perform XOR operation and append the result to the result string
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

// Example usage:
let result = stringXor(A: "010", b: "110")
print(result)  // Output: "100"
```

This function takes two strings A and B as input and returns the result of performing a bitwise XOR operation on each pair of characters. The strings must be of equal length.

func stringXor(a: String, b: String) -> String {
    // Ensure both strings are of equal length by padding the shorter one with zeros.
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation on each pair of characters.
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Ensure both strings are of equal length by padding the shorter one with zeros.
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation on each pair of characters.
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

// Example usage:
let result = stringXor(A: "010", b: "110")
print(result)  // Output: "100"
```

This function takes two strings representing binary numbers and performs a bitwise XOR operation on them. The XOR operation is defined as follows:
- If the bits at the same position in both numbers are the same, the result is 0.
- If the bits at the same position in both numbers are different, the result is 1.

The function ensures that the inputs are of equal length by padding the shorter one with leading zeros. The result of the XOR operation is returned as a string representing a binary number.

func stringXor(a: String, b: String) -> String {
    // Ensure the strings are of equal length by padding the shorter one with zeros
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Ensure the strings are of equal length by padding the shorter one with zeros
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

// Example usage:
let result = stringXor(A: "010", b: "110")
print(result)  // Output: "100"
```

This function takes two binary strings A and b, pads them to equal length, and performs a bitwise XOR operation on corresponding characters. The XOR operation is defined as follows:

- If the characters are the same, the result is '0'.
- If the characters are different, the result IS '1'.

The function returns the resulting binary string after performing the XOR operation.

func stringXor(a: String, b: String) -> String {
    // Check if the lengths of A and B are EQUAL
    guard A.count == b.count else {
        fatalError("Input strings must be of equal length.")
    }
    
    // Perform XOR operation on each pair of characters
    var result = ""
    for (charA, charB) in zip(A, b) {
        // Convert characters to integers (0 or 1)
        let intA = Int(String(charA))!
        let intB = Int(String(charB))!
        // Perform XOR and append to result string
        result.append(String((intA ^ intB) % 2))
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Check if the lengths of A and B are EQUAL
    guard A.count == b.count else {
        fatalError("Input strings must be of equal length.")
    }
    
    // Perform XOR operation on each pair of characters
    var result = ""
    for (charA, charB) in zip(A, b) {
        // Convert characters to integers (0 or 1)
        let intA = Int(String(charA))!
        let intB = Int(String(charB))!
        // Perform XOR and append to result string
        result.append(String((intA ^ intB) % 2))
    }
    
    return result
}

// Example usage:
let result = stringXor(A: "010", b: "110")
print(result)  // Output: "100"
```

This code defines a function `stringXor` that takes two strings `A` and `b` as input and returns their bitwise XOR as a string. The function first checks if the lengths of `A` and `b` are equal, and if not, it throws an error. It then iterates over each pair of characters, converts them to integers (0 or 1), performs XOR operation, and appends the result to the `result` string. Finally, it returns the `result` string.

func stringXor(a: String, b: String) -> String {
    // Check if the lengths of A and B are EQUAL
    if A.count != b.count {
        fatalError("Lengths of A and B must be equal")
    }
    
    // Initialize an empty string to store the result
    var result = ""
    
    // Perform XOR operation for each pair of characters
    for i in 0..<A.count {
        // Get the character at the current position in A and B
        let charA = String(A[A.index(A.startIndex, offsetBy: i)])
        let charB = String(b[b.index(b.startIndex, offsetBy: i)])
        
        // Perform XOR operation and append the result to the result string
        if charA == charB {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Check if the lengths of A and B are EQUAL
    if A.count != b.count {
        fatalError("Lengths of A and B must be equal")
    }
    
    // Initialize an empty string to store the result
    var result = ""
    
    // Perform XOR operation for each pair of characters
    for i in 0..<A.count {
        // Get the character at the current position in A and B
        let charA = String(A[A.index(A.startIndex, offsetBy: i)])
        let charB = String(b[b.index(b.startIndex, offsetBy: i)])
        
        // Perform XOR operation and append the result to the result string
        if charA == charB {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

// Example usage:
let result = stringXor(A: "010", b: "110")
print(result)  // Output: "100"
```

This function takes two strings A and B as input and returns the result of XOR operation on each pair of characters. The XOR operation is performed by comparing each character of the strings A and B at the same position and appending "0" if the characters are the same, and "1" if they are different.

func stringXor(a: String, b: String) -> String {
    // Ensure both strings are of equal length by padding the shorter one with zeros
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation on each pair of bits
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Ensure both strings are of equal length by padding the shorter one with zeros
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation on each pair of bits
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

// Example usage:
let result = stringXor(A: "010", b: "110")
print(result)  // Output: "100"
```

func stringXor(a: String, b: String) -> String {
    // Ensure the strings are of equal length by padding the shorter one with zeros.
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation by comparing each character.
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Ensure the strings are of equal length by padding the shorter one with zeros.
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation by comparing each character.
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

// Example usage:
let result = stringXor(A: "010", b: "110")
print(result)  // Output: "100"
```

This function takes two binary strings A and B, pads them to equal length, and performs a bitwise XOR operation on each pair of characters. The XOR operation is defined as follows:
- If the characters are the same, the result is '0'.
- If the characters are different, the result is '1'.

The function returns the resulting binary string after performing the XOR operation.

Note: The function assumes that the input strings contain only '1's and '0's and does not perform any validation. If other characters are present, the behavior is undefined.

func stringXor(a: String, b: String) -> String {
    // Ensure both strings are of equal length by padding the shorter one with zeros.
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation on each pair of characters.
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // Ensure both strings are of equal length by padding the shorter one with zeros.
    let maxLength = max(A.count, b.count)
    let paddedA = String(repeating: "0", count: maxLength - A.count) + A
    let paddedB = String(repeating: "0", count: maxLength - b.count) + b
    
    // Perform XOR operation on each pair of characters.
    var result = ""
    for i in 0..<maxLength {
        if paddedA[i] == paddedB[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

// Example usage:
let result = stringXor(A: "010", b: "110")
print(result)  // Output should be "100"
```

This function takes two binary strings A and b, pads the shorter one with leading zeros to match the length of the longer one, and then performs a bitwise XOR operation on each pair of characters from the two strings. The result is a new binary string where each bit is the result of XOR operation on the corresponding bits of the input strings.

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] != b[b.index(b.startIndex, offsetBy: i)] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] != b[b.index(b.startIndex, offsetBy: i)] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

func stringToInteger(input: String) -> Int {
    var result = 0
    for i in 0..<input.count {
        result = result * 10 + Int(input[input.index(input.startIndex, offsetBy: i)]) - 48
    }
    return result
}

func integerToString(input: Int) -> String {
    var result = ""
    var input = input
    while input > 0 {
        result = String(input % 10) + result
        input /= 10
    }
    return result
}

func sumArray(input: [Int]) -> Int {
    var result = 0
    for i in 0..<input.count {
        result += input[i]
    }
    return result
}

func productArray(input: [Int]) -> Int {
    var result = 1
    for i in 0..<input.count {
        result *= input[i]
    }
    return result
}

func averageArray(input: [Int]) -> Double {
    var result = 0.0
    for i in 0..<input.count {
        result += Double(input[i])
    }
    return result / Double(input.count)
}

func medianArray(input: [Int]) -> Double {
    var result = 0.0
    for i in 0..<input.count {
        result += Double(input[i])
    }
    return result / Double(input.count)
}

func modeArray(input: [Int]) -> Int {
    var result = 0
    for i in 0..<input.count {
        if input[i] > result {
            result = input[i]
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

// Input is a string of digits that represent a number.
// Return the number as an integer.
// stringToInteger(input: "123") == 123
func stringToInteger(input: String) -> Int {
    var result = 0
    for i in 0..<input.count {
        result = result * 10 + Int(input[input.index(input.startIndex, offsetBy: i)]) - 48
    }
    return result
}

// Input is a string of digits that represent a number.
// Return the number as an integer.
// integerToString(input: 123) == "123"
func integerToString(input: Int) -> String {
    var result = ""
    var input = input
    while input > 0 {
        result = String(input % 10) + result
        input /= 10
    }
    return result
}

// Input is an array of integers.
// Return the sum of all the integers in the array.
// sumArray(input: [1, 2, 3, 4]) == 10
func sumArray(input: [Int]) -> Int {
    var result = 0
    for i in 0..<input.count {
        result += input[i]
    }
    return result
}

// Input is an array of integers.
// Return the product of all the integers in the array.
// productArray(input: [1, 2, 3, 4]) == 24
func productArray(input: [Int]) -> Int {
    var result = 1
    for i in 0..<input.count {
        result *= input[i]
    }
    return result
}

// Input is an array of integers.
// Return the average of all the integers in the array.
// averageArray(input: [1, 2, 3, 4]) == 2.5
func averageArray(input: [Int]) -> Double {
    var result = 0.0
    for i in 0..<input.count {
        result += Double(input[i])
    }
    return result / Double(input.count)
}

// Input is an array of integers.
// Return the median of all the integers in the array.
// medianArray(input: [1, 2, 3, 4]) == 2.5
func medianArray(input: [Int]) -> Double {
    var result = 0.0
    for i in 0..<input.count {
        result += Double(input[i])
    }
    return result / Double(input.count)
}

// Input is an array of integers.
// Return the mode of all the integers in the array.
// modeArray(input: [1, 2, 3, 4]) == 1
func modeArray(input: [Int]) -> Int {
    var result = 0
    for i in 0..<input.count {
        if input[i] > result {
            result = input[i]
        }
    }
    return result
}

// Input is an array of integers.
// Return the mode of all the integers in the array.
// modeArray(input: [1, 2, 3, 4]) == 1
func modeArray(input: [Int]) -> Int {
    var result = 0
    for i in 0..<input.count {
        if input[i] > result {
            result = input[i]
        }
    }
    return result
}

// Input is an array of integers.
// Return the mode of all the integers in the array.
// modeArray(input: [1, 2, 3, 4]) == 1
func modeArray(input: [Int]) -> Int {
    var result = 0
    for i in 0..<input.count {
        if input[i] > result {
            result = input[i]
        }
    }
    return result
}

// Input is an array of integers

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] != b[b.index(b.startIndex, offsetBy: i)] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] != b[b.index(b.startIndex, offsetBy: i)] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] != b[b.index(b.startIndex, offsetBy: i)] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] != b[b.index(b.startIndex, offsetBy: i)] {
            result += "1"
        } else {
            result += "0"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
<|file_separator|>

func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    let c = zip(a, b).map { $0 ^ $1 }
    return String(c)
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let a = Array(a)
    let b = Array(b)
    let c = zip(a, b).map { $0 ^ $1 }
    return String(c)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result: [Character] = []
    for i in 0..<aArray.count {
        let aChar = aArray[i]
        let bChar = bArray[i]
        if aChar != bChar {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    return String(result)
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result: [Character] = []
    for i in 0..<aArray.count {
        let aChar = aArray[i]
        let bChar = bArray[i]
        if aChar != bChar {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
  return String(zip(a, b).map { $0 != $1 ? "1" : "0" }.joined())
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
  return String(zip(a, b).map { $0 != $1 ? "1" : "0" }.joined())
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    for i in 0..<a.count {
        if aChars[i] == bChars[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    for i in 0..<a.count {
        if aChars[i] == bChars[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
  var result = ""
  let aArray = Array(a)
  let bArray = Array(b)
  
  for (index, char) in aArray.enumerated() {
    if char == bArray[index] {
      result += "0"
    } else {
      result += "1"
    }
  }
  
  return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
  var result = ""
  let aArray = Array(a)
  let bArray = Array(b)
  
  for (index, char) in aArray.enumerated() {
    if char == bArray[index] {
      result += "0"
    } else {
      result += "1"
    }
  }
  
  return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aArray = a.map { String($0) }
    let bArray = b.map { String($0) }
    
    for i in 0 ..< aArray.count {
        if aArray[i] == bArray[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aArray = a.map { String($0) }
    let bArray = b.map { String($0) }
    
    for i in 0 ..< aArray.count {
        if aArray[i] == bArray[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
  let aArray = a.map { $0 }
  let bArray = b.map { $0 }
  var xorArray: [Character] = []
  for (index, _) in aArray.enumerated() {
    xorArray.append(aArray[index] == bArray[index] ? "0" : "1")
  }
  return String(xorArray)
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
  let aArray = a.map { $0 }
  let bArray = b.map { $0 }
  var xorArray: [Character] = []
  for (index, _) in aArray.enumerated() {
    xorArray.append(aArray[index] == bArray[index] ? "0" : "1")
  }
  return String(xorArray)
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        let charB = b[b.index(b.startIndex, offsetBy: index)]
        if char == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        let charB = b[b.index(b.startIndex, offsetBy: index)]
        if char == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
  // your code here
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
  // your code here
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        result += String(Int(String(char))! ^ Int(String(b[b.index(b.startIndex, offsetBy: index)]))!)
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        result += String(Int(String(char))! ^ Int(String(b[b.index(b.startIndex, offsetBy: index)]))!)
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = a.map(String.characterAtIndex)
    let b = b.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] ^ b[i]).toString()
    }
    return result
}

func stringAnd(a: String) -> String {
    let a = a.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] & a[i]).toString()
    }
    return result
}

func stringOr(a: String) -> String {
    let a = a.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] | a[i]).toString()
    }
    return result
}

func stringNot(a: String) -> String {
    let a = a.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] ^ 1).toString()
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let a = a.map(String.characterAtIndex)
    let b = b.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] ^ b[i]).toString()
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary AND on this input and return result also as a string.
// stringAnd(a: "110") == "110"
func stringAnd(a: String) -> String {
    let a = a.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] & a[i]).toString()
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary OR on this input and return result also as a string.
// stringOr(a: "110") == "111"
func stringOr(a: String) -> String {
    let a = a.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] | a[i]).toString()
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary NOT on this input and return result also as a string.
// stringNot(a: "110") == "001"
func stringNot(a: String) -> String {
    let a = a.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] ^ 1).toString()
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on this input and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let a = a.map(String.characterAtIndex)
    let b = b.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] ^ b[i]).toString()
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary AND on this input and return result also as a string.
// stringAnd(a: "110") == "110"
func stringAnd(a: String) -> String {
    let a = a.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] & a[i]).toString()
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary OR on this input and return result also as a string.
// stringOr(a: "110") == "111"
func stringOr(a: String) -> String {
    let a = a.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] | a[i]).toString()
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary NOT on this input and return result also as a string.
// stringNot(a: "110") == "001"
func stringNot(a: String) -> String {
    let a = a.map(String.characterAtIndex)
    var result = ""
    for i in 0..<a.count {
        result += (a[i] ^ 1).toString()
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on this input and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let a = a.map(String.characterAtIndex)
    let b = b.map(String

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.length, b.length) {
        result += (a[i] ^ b[i]).toString()
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.length, b.length) {
        result += (a[i] ^ b[i]).toString()
    }
    return result
}

// Test cases
print(stringXor(a: "010", b: "110")) // Output: "100"
print(stringXor(a: "101", b: "101")) // Output: "000"
print(stringXor(a: "111", b: "000")) // Output: "111"
print(stringXor(a: "000", b: "000")) // Output: "000"
print(stringXor(a: "1111", b: "1111")) // Output: "0000"
print(stringXor(a: "1111", b: "11111")) // Output: "000000"
```

func stringXor(a: String, b: String) -> String {
    let result = "";
    for i in 0..<max(a.length, b.length) {
        result += (a.characterAtIndex(i) ^ b.characterAtIndex(i)).stringValue;
    }
    return result;
}

func stringXor(a: String) -> String {
    let result = "";
    for i in 0..<a.length {
        result += (a.characterAtIndex(i) ^ '0').stringValue;
    }
    return result;
}

func stringXor(a: String) -> String {
    let result = "";
    for i in 0..<a.length {
        result += (a.characterAtIndex(i) ^ '1').stringValue;
    }
    return result;
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let result = "";
    for i in 0..<max(a.length, b.length) {
        result += (a.characterAtIndex(i) ^ b.characterAtIndex(i)).stringValue;
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on the input string and return result also as a string.
// stringXor(a: "010") == "101"
func stringXor(a: String) -> String {
    let result = "";
    for i in 0..<a.length {
        result += (a.characterAtIndex(i) ^ '0').stringValue;
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on the input string and return result also as a string.
// stringXor(a: "110") == "001"
func stringXor(a: String) -> String {
    let result = "";
    for i in 0..<a.length {
        result += (a.characterAtIndex(i) ^ '1').stringValue;
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on the input string and return result also as a string.
// stringXor(a: "111") == "000"
func stringXor(a: String) -> String {
    let result = "";
    for i in 0..<a.length {
        result += (a.characterAtIndex(i) ^ '0').stringValue;
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on the input string and return result also as a string.
// stringXor(a: "000") == "111"
func stringXor(a: String) -> String {
    let result = "";
    for i in 0..<a.length {
        result += (a.characterAtIndex(i) ^ '1').stringValue;
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on the input string and return result also as a string.
// stringXor(a: "001") == "110"
func stringXor(a: String) -> String {
    let result = "";
    for i in 0..<a.length {
        result += (a.characterAtIndex(i) ^ '0').stringValue;
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on the input string and return result also as a string.
// stringXor(a: "011") == "100"
func stringXor(a: String) -> String {
    let result = "";
    for i in 0..<a.length {
        result += (a.characterAtIndex(i) ^ '1').stringValue;
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on the input string and return result also as a string.
// stringXor(a: "101") == "010"
func stringXor(a: String) -> String {
    let result = "";
    for i in 0..<a.length {
        result += (a.characterAtIndex(i) ^ '0').stringValue;
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on the input string and return result also as a string.
// stringXor(a: "100") == "011"
func stringXor(a: String) -> String {
    let result = "";
    for i in 0..<a.length {
        result += (a.characterAtIndex(i) ^ '1').stringValue;
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on the input string and return result also as a string.
// stringXor(a: "010") == "101"
func stringXor(a: String) -> String {
    let result = "";
    for i in 0..<a.length {
        result += (a.characterAtIndex(i) ^ '

func stringXor(a: String, b: String) -> String {
    let result = "";
    for i in 0..<max(a.length, b.length) {
        let digitA = if i < a.length { a[i] } else { '0' };
        let digitB = if i < b.length { b[i] } else { '0' };
        result += (digitA ^ digitB).toString();
    }
    return result;
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    let result = "";
    for i in 0..<max(a.length, b.length) {
        let digitA = if i < a.length { a[i] } else { '0' };
        let digitB = if i < b.length { b[i] } else { '0' };
        result += (digitA ^ digitB).toString();
    }
    return result;
}

// Test the function
let a = "010";
let b = "110";
let result = stringXor(a, b);
print(result); // Output: "100"
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        result += (a.characterAtIndex(i) ^ b.characterAtIndex(i)).stringValue
    }
    return result
}

func stringAnd(a: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) & a.characterAtIndex(i)).stringValue
    }
    return result
}

func stringOr(a: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) | a.characterAtIndex(i)).stringValue
    }
    return result
}

func stringNot(a: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) ^ 1).stringValue
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        result += (a.characterAtIndex(i) ^ b.characterAtIndex(i)).stringValue
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary AND on this input and return result also as a string.
// stringAnd(a: "010") == "010"
func stringAnd(a: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) & a.characterAtIndex(i)).stringValue
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary OR on this input and return result also as a string.
// stringOr(a: "010") == "110"
func stringOr(a: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) | a.characterAtIndex(i)).stringValue
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary NOT on this input and return result also as a string.
// stringNot(a: "010") == "101"
func stringNot(a: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) ^ 1).stringValue
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on this input and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        result += (a.characterAtIndex(i) ^ b.characterAtIndex(i)).stringValue
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary AND on this input and return result also as a string.
// stringAnd(a: "010") == "010"
func stringAnd(a: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) & a.characterAtIndex(i)).stringValue
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary OR on this input and return result also as a string.
// stringOr(a: "010") == "110"
func stringOr(a: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) | a.characterAtIndex(i)).stringValue
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary NOT on this input and return result also as a string.
// stringNot(a: "010") == "101"
func stringNot(a: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) ^ 1).stringValue
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on this input and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.count, b.count) {
        result += (a.characterAtIndex(i) ^ b.characterAtIndex(i)).stringValue
    }
    return result
}

// Input is a string consisting of 1s and 0s.
// Perform binary AND on this input and return result also as a string.
// stringAnd(a: "010") == "010"
func stringAnd(a: String) -> String

func stringXor(a: String, b: String) -> String {
    var result = "";
    for i in 0..<max(a.count, b.count) {
        result += (a.characterAtIndex(i) ^ b.characterAtIndex(i)).stringValue;
    }
    return result;
}

func arrayXor(a: [Int]) -> [Int] {
    var result = [Int](nil);
    for i in 0..<a.count {
        result.append(a[i] ^ a[i + 1]);
    }
    return result;
}

func stringXor(a: String) -> String {
    var result = "";
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) ^ a.characterAtIndex(i + 1)).stringValue;
    }
    return result;
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = "";
    for i in 0..<max(a.count, b.count) {
        result += (a.characterAtIndex(i) ^ b.characterAtIndex(i)).stringValue;
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all elements of the array and return result also as an array.
// arrayXor([1, 2, 3]) == [4, 0, 7]
func arrayXor(a: [Int]) -> [Int] {
    var result = [Int](nil);
    for i in 0..<a.count {
        result.append(a[i] ^ a[i + 1]);
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on all elements of the string and return result also as a string.
// stringXor("010") == "110"
func stringXor(a: String) -> String {
    var result = "";
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) ^ a.characterAtIndex(i + 1)).stringValue;
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on all elements of the string and return result also as a string.
// stringXor("010") == "110"
func stringXor(a: String) -> String {
    var result = "";
    for i in 0..<a.count {
        result += (a.characterAtIndex(i) ^ a.characterAtIndex(i + 1)).stringValue;
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all elements of the array and return result also as an array.
// arrayXor([1, 2, 3]) == [4, 0, 7]
func arrayXor(a: [Int]) -> [Int] {
    var result = [Int](nil);
    for i in 0..<a.count {
        result.append(a[i] ^ a[i + 1]);
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all elements of the array and return result also as an array.
// arrayXor([1, 2, 3]) == [4, 0, 7]
func arrayXor(a: [Int]) -> [Int] {
    var result = [Int](nil);
    for i in 0..<a.count {
        result.append(a[i] ^ a[i + 1]);
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all elements of the array and return result also as an array.
// arrayXor([1, 2, 3]) == [4, 0, 7]
func arrayXor(a: [Int]) -> [Int] {
    var result = [Int](nil);
    for i in 0..<a.count {
        result.append(a[i] ^ a[i + 1]);
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all elements of the array and return result also as an array.
// arrayXor([1, 2, 3]) == [4, 0, 7]
func arrayXor(a: [Int]) -> [Int] {
    var result = [Int](nil);
    for i in 0..<a.count {
        result.append(a[i] ^ a[i + 1]);
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all elements of the array and return result also as an array.
// arrayXor([1, 2, 3]) == [4, 0, 7]
func arrayXor(a: [Int]) -> [Int] {
    var result = [Int](nil);
    for i in 0..<a.count {
        result.append(a[i] ^ a[i + 1]);
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all elements of the array and return result also as an array.
// arrayXor([1, 2, 3])

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.length, b.length) {
        var aChar = if i < a.length { a[i] } else { '0' }
        var bChar = if i < b.length { b[i] } else { '0' }
        result += String(aChar ^ bChar)
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.length, b.length) {
        var aChar = if i < a.length { a[i] } else { '0' }
        var bChar = if i < b.length { b[i] } else { '0' }
        result += String(aChar ^ bChar)
    }
    return result
}

// Test cases
print(stringXor(a: "010", b: "110")) // Output: "100"
print(stringXor(a: "101", b: "011")) // Output: "110"
print(stringXor(a: "10010", b: "10101")) // Output: "00111"
print(stringXor(a: "11100", b: "00011")) // Output: "11111"
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.length, b.length) {
        if i < a.length && i < b.length {
            result += (a[i] ^ b[i]).toString()
        } else if i < a.length {
            result += a[i].toString()
        } else {
            result += b[i].toString()
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<max(a.length, b.length) {
        if i < a.length && i < b.length {
            result += (a[i] ^ b[i]).toString()
        } else if i < a.length {
            result += a[i].toString()
        } else {
            result += b[i].toString()
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = "";
    for i in 0..<max(a.length, b.length) {
        result += (a[i] ^ b[i]).toString();
    }
    return result;
}

func arrayXor(arr: [Int]) -> String {
    var result = "";
    for i in 0..<arr.count {
        result += (arr[i] ^ arr[i + 1]).toString();
    }
    return result;
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = "";
    for i in 0..<max(a.length, b.length) {
        result += (a[i] ^ b[i]).toString();
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all the elements of the array and return the result as a string.
// arrayXor([1, 2, 3, 4]) == "11"
func arrayXor(arr: [Int]) -> String {
    var result = "";
    for i in 0..<arr.count {
        result += (arr[i] ^ arr[i + 1]).toString();
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all the elements of the array and return the result as a string.
// arrayXor([1, 2, 3, 4]) == "11"
func arrayXor(arr: [Int]) -> String {
    var result = "";
    for i in 0..<arr.count {
        result += (arr[i] ^ arr[i + 1]).toString();
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all the elements of the array and return the result as a string.
// arrayXor([1, 2, 3, 4]) == "11"
func arrayXor(arr: [Int]) -> String {
    var result = "";
    for i in 0..<arr.count {
        result += (arr[i] ^ arr[i + 1]).toString();
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all the elements of the array and return the result as a string.
// arrayXor([1, 2, 3, 4]) == "11"
func arrayXor(arr: [Int]) -> String {
    var result = "";
    for i in 0..<arr.count {
        result += (arr[i] ^ arr[i + 1]).toString();
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all the elements of the array and return the result as a string.
// arrayXor([1, 2, 3, 4]) == "11"
func arrayXor(arr: [Int]) -> String {
    var result = "";
    for i in 0..<arr.count {
        result += (arr[i] ^ arr[i + 1]).toString();
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all the elements of the array and return the result as a string.
// arrayXor([1, 2, 3, 4]) == "11"
func arrayXor(arr: [Int]) -> String {
    var result = "";
    for i in 0..<arr.count {
        result += (arr[i] ^ arr[i + 1]).toString();
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all the elements of the array and return the result as a string.
// arrayXor([1, 2, 3, 4]) == "11"
func arrayXor(arr: [Int]) -> String {
    var result = "";
    for i in 0..<arr.count {
        result += (arr[i] ^ arr[i + 1]).toString();
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all the elements of the array and return the result as a string.
// arrayXor([1, 2, 3, 4]) == "11"
func arrayXor(arr: [Int]) -> String {
    var result = "";
    for i in 0..<arr.count {
        result += (arr[i] ^ arr[i + 1]).toString();
    }
    return result;
}

// Input is an array of integers.
// Perform binary XOR on all the elements of the array and return the result as a string.
// arrayXor([1, 2, 3, 4]) == "11"
func arrayXor(arr: [Int]) -> String {
    var result = "";
    for i in 0..<arr.count {
        result += (arr[i] ^ arr[

func stringXor(a: String, b: String) -> String {
    var result = "";
    for i in 0..<max(a.length, b.length) {
        if a.length > i {
            result += a[i];
        }
        if b.length > i {
            result += b[i];
        }
    }
    return result;
}

func stringAnd(a: String) -> String {
    var result = "";
    for i in 0..<a.length {
        if a[i] == '1' {
            result += a[i];
        }
    }
    return result;
}

func stringOr(a: String) -> String {
    var result = "";
    for i in 0..<a.length {
        if a[i] == '1' {
            result += a[i];
        }
    }
    return result;
}

func stringNot(a: String) -> String {
    var result = "";
    for i in 0..<a.length {
        if a[i] == '1' {
            result += '0';
        } else {
            result += '1';
        }
    }
    return result;
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = "";
    for i in 0..<max(a.length, b.length) {
        if a.length > i {
            result += a[i];
        }
        if b.length > i {
            result += b[i];
        }
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary AND on this input and return result also as a string.
// stringAnd(a: "010") == "010"
func stringAnd(a: String) -> String {
    var result = "";
    for i in 0..<a.length {
        if a[i] == '1' {
            result += a[i];
        }
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary OR on this input and return result also as a string.
// stringOr(a: "010") == "011"
func stringOr(a: String) -> String {
    var result = "";
    for i in 0..<a.length {
        if a[i] == '1' {
            result += a[i];
        }
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary NOT on this input and return result also as a string.
// stringNot(a: "010") == "101"
func stringNot(a: String) -> String {
    var result = "";
    for i in 0..<a.length {
        if a[i] == '1' {
            result += '0';
        } else {
            result += '1';
        }
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on this input and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = "";
    for i in 0..<max(a.length, b.length) {
        if a.length > i {
            result += a[i];
        }
        if b.length > i {
            result += b[i];
        }
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary AND on this input and return result also as a string.
// stringAnd(a: "010") == "010"
func stringAnd(a: String) -> String {
    var result = "";
    for i in 0..<a.length {
        if a[i] == '1' {
            result += a[i];
        }
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary OR on this input and return result also as a string.
// stringOr(a: "010") == "011"
func stringOr(a: String) -> String {
    var result = "";
    for i in 0..<a.length {
        if a[i] == '1' {
            result += a[i];
        }
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary NOT on this input and return result also as a string.
// stringNot(a: "010") == "101"
func stringNot(a: String) -> String {
    var result = "";
    for i in 0..<a.length {
        if a[i] == '1' {
            result += '0';
        } else {
            result += '1';
        }
    }
    return result;
}

// Input is a string consisting of 1s and 0s.
// Perform binary XOR on this input and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = "";
    for i in 0..<max(a.length, b.length) {
        if

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aBit = Int(a[a.index(a.startIndex, offsetBy: i)])!
        let bBit = Int(b[b.index(b.startIndex, offsetBy: i)])!
        result.append(Int(aBit ^ bBit)!)
    }
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aBit = Int(a[a.index(a.startIndex, offsetBy: i)])!
        let bBit = Int(b[b.index(b.startIndex, offsetBy: i)])!
        result.append(Int(aBit ^ bBit)!)
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] ^ b[i])
    }
    return result
}

extension String {
    subscript(i: Int) -> Character {
        return self[index(startIndex, offsetBy: i)]
    }
    
    func ^ (lhs: String, rhs: String) -> Character {
        return lhs == rhs ? "0" : "1"
    }
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] ^ b[i])
    }
    return result
}

extension String {
    subscript(i: Int) -> Character {
        return self[index(startIndex, offsetBy: i)]
    }
    
    func ^ (lhs: String, rhs: String) -> Character {
        return lhs == rhs ? "0" : "1"
    }
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] ^ b[i])
    }
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] ^ b[i])
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        result.append(aChar ^ bChar)
    }
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        result.append(aChar ^ bChar)
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] == b[i] ? "0" : "1")
    }
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] == b[i] ? "0" : "1")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] ^ b[i])
    }
    return result
}

extension String {
    subscript(i: Int) -> Character {
        return self[index(at: i)]
    }

    func ^ (lhs: Character, rhs: Character) -> Character {
        return lhs == rhs ? "0" : "1"
    }
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        result.append(a[i] ^ b[i])
    }
    return result
}

extension String {
    subscript(i: Int) -> Character {
        return self[index(at: i)]
    }

    func ^ (lhs: Character, rhs: Character) -> Character {
        return lhs == rhs ? "0" : "1"
    }
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = a[a.index(a.startIndex, offsetBy: i)]
        let bBit = b[b.index(b.startIndex, offsetBy: i)]
        
        result.append(aBit == bBit ? "0" : "1")
    }
    
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = a[a.index(a.startIndex, offsetBy: i)]
        let bBit = b[b.index(b.startIndex, offsetBy: i)]
        
        result.append(aBit == bBit ? "0" : "1")
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = a[a.index(a.startIndex, offsetBy: i)]
        let bBit = b[b.index(b.startIndex, offsetBy: i)]
        
        result.append(aBit == bBit ? "0" : "1")
    }
    
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    
    for i in 0..<a.count {
        let aBit = a[a.index(a.startIndex, offsetBy: i)]
        let bBit = b[b.index(b.startIndex, offsetBy: i)]
        
        result.append(aBit == bBit ? "0" : "1")
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    
    return result
}

<start_of_turn>user
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"<end_of_turn>
<start_of_turn>model
```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    let aChars = Array(a)
    let bChars = Array(b)
    
    for i in 0..<aChars.count {
        result.append(aChars[i] == bChars[i] ? "0" : "1")
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = a.binary
    let bBinary = b.binary
    let result = aBinary ^ bBinary
    return result.binary
}

extension String {
    var binary: String {
        return self.map { String($0, radix: 2) }.joined()
    }
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = a.binary
    let bBinary = b.binary
    let result = aBinary ^ bBinary
    return result.binary
}

extension String {
    var binary: String {
        return self.map { String($0, radix: 2) }.joined()
    }
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = a.map { Int(String($0))! }
    let bArray = b.map { Int(String($0))! }
    let resultArray = zip(aArray, bArray).map { $0 ^ $1 }
    return resultArray.map { String($0) }.joined()
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aArray = a.map { Int(String($0))! }
    let bArray = b.map { Int(String($0))! }
    let resultArray = zip(aArray, bArray).map { $0 ^ $1 }
    return resultArray.map { String($0) }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aBits = a.map { Int(String($0))! }
    let bBits = b.map { Int(String($0))! }
    let resultBits = zip(aBits, bBits).map { $0 ^ $1 }
    return resultBits.map { String($0) }.joined()
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aBits = a.map { Int(String($0))! }
    let bBits = b.map { Int(String($0))! }
    let resultBits = zip(aBits, bBits).map { $0 ^ $1 }
    return resultBits.map { String($0) }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = a.map { Int(String($0))! }
    let bArray = b.map { Int(String($0))! }
    let resultArray = zip(aArray, bArray).map { $0 ^ $1 }
    return resultArray.map { String($0) }.joined()
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aArray = a.map { Int(String($0))! }
    let bArray = b.map { Int(String($0))! }
    let resultArray = zip(aArray, bArray).map { $0 ^ $1 }
    return resultArray.map { String($0) }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = a.map { Int(String($0))! }
    let bArray = b.map { Int(String($0))! }
    let resultArray = zip(aArray, bArray).map { $0 ^ $1 }
    return resultArray.map { String($0) }.joined()
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aArray = a.map { Int(String($0))! }
    let bArray = b.map { Int(String($0))! }
    let resultArray = zip(aArray, bArray).map { $0 ^ $1 }
    return resultArray.map { String($0) }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = a.binary
    let bBinary = b.binary
    let result = aBinary ^ bBinary
    return result.binary
}

extension String {
    var binary: String {
        return self.reduce("") { $0 + String($1, radix: 2) }
    }
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = a.binary
    let bBinary = b.binary
    let result = aBinary ^ bBinary
    return result.binary
}

extension String {
    var binary: String {
        return self.reduce("") { $0 + String($1, radix: 2) }
    }
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = a.map { Int(String($0))! }
    let bArray = b.map { Int(String($0))! }
    let resultArray = zip(aArray, bArray).map { $0 ^ $1 }
    return resultArray.map { String($0) }.joined()
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aArray = a.map { Int(String($0))! }
    let bArray = b.map { Int(String($0))! }
    let resultArray = zip(aArray, bArray).map { $0 ^ $1 }
    return resultArray.map { String($0) }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = a.map { Int(String($0))! }
    let bBinary = b.map { Int(String($0))! }
    let result = zip(aBinary, bBinary).map { $0 ^ $1 }
    return result.map { String($0) }.joined()
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = a.map { Int(String($0))! }
    let bBinary = b.map { Int(String($0))! }
    let result = zip(aBinary, bBinary).map { $0 ^ $1 }
    return result.map { String($0) }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = a.map { $0 == "1" ? 1 : 0 }
    let bBinary = b.map { $0 == "1" ? 1 : 0 }
    let result = zip(aBinary, bBinary).map { $0 ^ $1 }
    return result.map { $0 == 1 ? "1" : "0" }.joined()
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = a.map { $0 == "1" ? 1 : 0 }
    let bBinary = b.map { $0 == "1" ? 1 : 0 }
    let result = zip(aBinary, bBinary).map { $0 ^ $1 }
    return result.map { $0 == 1 ? "1" : "0" }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = a.map { Int(String($0))! }
    let bArray = b.map { Int(String($0))! }
    let resultArray = zip(aArray, bArray).map { $0 ^ $1 }
    return resultArray.map { String($0) }.joined()
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aArray = a.map { Int(String($0))! }
    let bArray = b.map { Int(String($0))! }
    let resultArray = zip(aArray, bArray).map { $0 ^ $1 }
    return resultArray.map { String($0) }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        if char == "1" {
            result += b[i] == "1" ? "0" : "1"
        } else {
            result += b[i] == "0" ? "0" : "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        if char == "1" {
            result += b[i] == "1" ? "0" : "1"
        } else {
            result += b[i] == "0" ? "0" : "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = a.binaryValue
    let bBinary = b.binaryValue
    let result = aBinary ^ bBinary
    return result.binaryString
}

extension String {
    var binaryValue: Int {
        return Int(self, radix: 2)!
    }

    var binaryString: String {
        return String(self, radix: 2)
    }
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = a.binaryValue
    let bBinary = b.binaryValue
    let result = aBinary ^ bBinary
    return result.binaryString
}

extension String {
    var binaryValue: Int {
        return Int(self, radix: 2)!
    }

    var binaryString: String {
        return String(self, radix: 2)
    }
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        let bChar = b[i]
        if char == "1" && bChar == "1" {
            result += "1"
        } else if char == "0" && bChar == "0" {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        let bChar = b[i]
        if char == "1" && bChar == "1" {
            result += "1"
        } else if char == "0" && bChar == "0" {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" ? 1 : 0 }
    let b = b.map { $0 == "1" ? 1 : 0 }
    let result = zip(a, b).map { $0 ^ $1 }
    return result.map { $0 == 1 ? "1" : "0" }.joined()
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" ? 1 : 0 }
    let b = b.map { $0 == "1" ? 1 : 0 }
    let result = zip(a, b).map { $0 ^ $1 }
    return result.map { $0 == 1 ? "1" : "0" }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        if char == "1" {
            result += b[i] == "1" ? "0" : "1"
        } else {
            result += b[i] == "0" ? "0" : "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        if char == "1" {
            result += b[i] == "1" ? "0" : "1"
        } else {
            result += b[i] == "0" ? "0" : "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, c) in a.enumerated() {
        let d = b[i]
        if c == "1" && d == "0" {
            result.append("1")
        } else if c == "0" && d == "1" {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, c) in a.enumerated() {
        let d = b[i]
        if c == "1" && d == "0" {
            result.append("1")
        } else if c == "0" && d == "1" {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, c) in a.enumerated() {
        let d = b[i]
        if c == "1" && d == "1" {
            result.append("0")
        } else if c == "0" && d == "0" {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, c) in a.enumerated() {
        let d = b[i]
        if c == "1" && d == "1" {
            result.append("0")
        } else if c == "0" && d == "0" {
            result.append("1")
        } else {
            result.append("0")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aBinary = a.binaryValue
    let bBinary = b.binaryValue
    let result = aBinary ^ bBinary
    return result.binaryString
}

extension String {
    var binaryValue: Int {
        return Int(self, radix: 2) ?? 0
    }
}

extension Int {
    var binaryString: String {
        return String(self, radix: 2)
    }
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aBinary = a.binaryValue
    let bBinary = b.binaryValue
    let result = aBinary ^ bBinary
    return result.binaryString
}

extension String {
    var binaryValue: Int {
        return Int(self, radix: 2) ?? 0
    }
}

extension Int {
    var binaryString: String {
        return String(self, radix: 2)
    }
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        if char == "1" {
            result += b[i] == "1" ? "0" : "1"
        } else {
            result += b[i] == "0" ? "0" : "1"
        }
    }
    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        if char == "1" {
            result += b[i] == "1" ? "0" : "1"
        } else {
            result += b[i] == "0" ? "0" : "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aChars = Array(a)
    let bChars = Array(b)
    var result = ""

    for i in 0..<aChars.count {
        if aChars[i] == "1" && bChars[i] == "1" {
            result += "1"
        } else {
            result += "0"
        }
    }

    return result
}

[INST] Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100" [/INST] ```swift
func stringXor(a: String, b: String) -> String {
    let aChars = Array(a)
    let bChars = Array(b)
    var result = ""

    for i in 0..<aChars.count {
        if aChars[i] == "1" && bChars[i] == "1" {
            result += "1"
        } else {
            result += "0"
        }
    }

    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aAsBits = Array(a.reversed())
    let bAsBits = Array(b.reversed())
    
    var result = ""
    
    for i in 0..<max(aAsBits.count, bAsBits.count) {
        let aBit = aAsBits.count > i? aAsBits[i] : "0"
        let bBit = bAsBits.count > i? bAsBits[i] : "0"
        
        let xorBit = aBit == "1"? (bBit == "1"? "0" : "1") : (bBit == "1"? "1" : "0")
        
        result = "\(xorBit)\(result)"
    }
    
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aAsBits = Array(a.reversed())
    let bAsBits = Array(b.reversed())
    
    var result = ""
    
    for i in 0..<max(aAsBits.count, bAsBits.count) {
        let aBit = aAsBits.count > i? aAsBits[i] : "0"
        let bBit = bAsBits.count > i? bAsBits[i] : "0"
        
        let xorBit = aBit == "1"? (bBit == "1"? "0" : "1") : (bBit == "1"? "1" : "0")
        
        result = "\(xorBit)\(result)"
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aAsBits = Array(a.reversed())
    let bAsBits = Array(b.reversed())
    
    var result = ""
    
    for i in 0..<max(aAsBits.count, bAsBits.count) {
        let aBit = aAsBits.count > i? aAsBits[i] : "0"
        let bBit = bAsBits.count > i? bAsBits[i] : "0"
        
        let xorBit = (aBit == "1" && bBit == "0") || (aBit == "0" && bBit == "1")? "1" : "0"
        
        result = "\(xorBit)\(result)"
    }
    
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aAsBits = Array(a.reversed())
    let bAsBits = Array(b.reversed())
    
    var result = ""
    
    for i in 0..<max(aAsBits.count, bAsBits.count) {
        let aBit = aAsBits.count > i? aAsBits[i] : "0"
        let bBit = bAsBits.count > i? bAsBits[i] : "0"
        
        let xorBit = (aBit == "1" && bBit == "0") || (aBit == "0" && bBit == "1")? "1" : "0"
        
        result = "\(xorBit)\(result)"
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aAsBits = Array(a.reversed())
    let bAsBits = Array(b.reversed())
    
    var result = Array(repeating: 0, count: max(aAsBits.count, bAsBits.count))
    
    for i in 0..<result.count {
        let aBit = aAsBits.count > i? aAsBits[i] : 0
        let bBit = bAsBits.count > i? bAsBits[i] : 0
        
        result[i] = aBit ^ bBit
    }
    
    return String(result.reversed())
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aAsBits = Array(a.reversed())
    let bAsBits = Array(b.reversed())
    
    var result = Array(repeating: 0, count: max(aAsBits.count, bAsBits.count))
    
    for i in 0..<result.count {
        let aBit = aAsBits.count > i? aAsBits[i] : 0
        let bBit = bAsBits.count > i? bAsBits[i] : 0
        
        result[i] = aBit ^ bBit
    }
    
    return String(result.reversed())
}
```

func stringXor(a: String, b: String) -> String {
    let aAsBits = Array(a.reversed())
    let bAsBits = Array(b.reversed())
    
    var resultBits = Array(repeating: 0, count: max(aAsBits.count, bAsBits.count))
    
    for i in 0..<resultBits.count {
        if aAsBits.count > i {
            resultBits[i] ^= aAsBits[i]
        }
        if bAsBits.count > i {
            resultBits[i] ^= bAsBits[i]
        }
    }
    
    return String(resultBits.reversed())
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aAsBits = Array(a.reversed())
    let bAsBits = Array(b.reversed())
    
    var resultBits = Array(repeating: 0, count: max(aAsBits.count, bAsBits.count))
    
    for i in 0..<resultBits.count {
        if aAsBits.count > i {
            resultBits[i] ^= aAsBits[i]
        }
        if bAsBits.count > i {
            resultBits[i] ^= bAsBits[i]
        }
    }
    
    return String(resultBits.reversed())
}
```

func stringXor(a: String, b: String) -> String {
    guard a.count == b.count else {
        return ""
    }
    
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        let xorResult = Int(String(aChar))! ^ Int(String(bChar))!
        result += String(xorResult)
    }
    
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    guard a.count == b.count else {
        return ""
    }
    
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        let xorResult = Int(String(aChar))! ^ Int(String(bChar))!
        result += String(xorResult)
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aAsBits = Array(a.reversed())
    let bAsBits = Array(b.reversed())
    
    var result = ""
    
    for i in 0..<max(aAsBits.count, bAsBits.count) {
        let aBit = aAsBits.count > i? aAsBits[i] : "0"
        let bBit = bAsBits.count > i? bAsBits[i] : "0"
        
        let xorBit = (aBit == "1" && bBit == "0") || (aBit == "0" && bBit == "1")? "1" : "0"
        
        result = "\(xorBit)\(result)"
    }
    
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aAsBits = Array(a.reversed())
    let bAsBits = Array(b.reversed())
    
    var result = ""
    
    for i in 0..<max(aAsBits.count, bAsBits.count) {
        let aBit = aAsBits.count > i? aAsBits[i] : "0"
        let bBit = bAsBits.count > i? bAsBits[i] : "0"
        
        let xorBit = (aBit == "1" && bBit == "0") || (aBit == "0" && bBit == "1")? "1" : "0"
        
        result = "\(xorBit)\(result)"
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aInt = Int(a, radix: 2)!
    let bInt = Int(b, radix: 2)!
    let resultInt = aInt ^ bInt
    return String(resultInt, radix: 2)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aInt = Int(a, radix: 2)!
    let bInt = Int(b, radix: 2)!
    let resultInt = aInt ^ bInt
    return String(resultInt, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    guard a.count == b.count else {
        return ""
    }
    
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        let xorChar = Int(String(aChar))! ^ Int(String(bChar))!
        result += String(xorChar)
    }
    
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    guard a.count == b.count else {
        return ""
    }
    
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        let xorChar = Int(String(aChar))! ^ Int(String(bChar))!
        result += String(xorChar)
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    guard a.count == b.count else {
        return ""
    }
    
    var result = ""
    for i in 0..<a.count {
        let bitA = a[a.index(a.startIndex, offsetBy: i)]
        let bitB = b[b.index(b.startIndex, offsetBy: i)]
        
        if bitA == "1" && bitB == "1" {
            result.append("0")
        } else if bitA == "0" && bitB == "0" {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    guard a.count == b.count else {
        return ""
    }
    
    var result = ""
    for i in 0..<a.count {
        let bitA = a[a.index(a.startIndex, offsetBy: i)]
        let bitB = b[b.index(b.startIndex, offsetBy: i)]
        
        if bitA == "1" && bitB == "1" {
            result.append("0")
        } else if bitA == "0" && bitB == "0" {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    guard a.count == b.count else {
        return ""
    }
    
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        let xorResult = Int(String(aChar))! ^ Int(String(bChar))!
        result += String(xorResult)
    }
    
    return result
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    guard a.count == b.count else {
        return ""
    }
    
    var result = ""
    for i in 0..<a.count {
        let aChar = a[a.index(a.startIndex, offsetBy: i)]
        let bChar = b[b.index(b.startIndex, offsetBy: i)]
        
        let xorResult = Int(String(aChar))! ^ Int(String(bChar))!
        result += String(xorResult)
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))?? 0
        let bInt = Int(String(bArray[i]))?? 0
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))?? 0
        let bInt = Int(String(bArray[i]))?? 0
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))?? 0
        let bInt = Int(String(bArray[i]))?? 0
        let xorInt = aInt ^ bInt
        let xorStr = String(xorInt)
        result.append(Character(xorStr))
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))?? 0
        let bInt = Int(String(bArray[i]))?? 0
        let xorInt = aInt ^ bInt
        let xorStr = String(xorInt)
        result.append(Character(xorStr))
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))?? 0
        let bInt = Int(String(bArray[i]))?? 0
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))?? 0
        let bInt = Int(String(bArray[i]))?? 0
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()

    for (index, _) in aArray.enumerated() {
        let aBit = aArray[index] == "1"? 1 : 0
        let bBit = bArray[index] == "1"? 1 : 0
        let xorBit = aBit ^ bBit
        let xorChar = xorBit == 1? "1" : "0"
        result.append(Character(xorChar))
    }

    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()

    for (index, _) in aArray.enumerated() {
        let aBit = aArray[index] == "1"? 1 : 0
        let bBit = bArray[index] == "1"? 1 : 0
        let xorBit = aBit ^ bBit
        let xorChar = xorBit == 1? "1" : "0"
        result.append(Character(xorChar))
    }

    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var resultArray = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        resultArray.append(xorChar)
    }
    
    return String(resultArray)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var resultArray = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        resultArray.append(xorChar)
    }
    
    return String(resultArray)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorResult = aInt ^ bInt
        result.append(Character(UnicodeScalar(xorResult)!))
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorResult = aInt ^ bInt
        result.append(Character(UnicodeScalar(xorResult)!))
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for (i, char) in aArray.enumerated() {
        let aInt = Int(String(char))!
        let bInt = Int(String(bArray[i]))!
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for (i, char) in aArray.enumerated() {
        let aInt = Int(String(char))!
        let bInt = Int(String(bArray[i]))!
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}

Question:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"

Answer:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    var result = [Character]()
    
    for i in 0..<aArray.count {
        let aInt = Int(String(aArray[i]))!
        let bInt = Int(String(bArray[i]))!
        let xorInt = aInt ^ bInt
        let xorChar = Character(UnicodeScalar(xorInt)!)
        result.append(xorChar)
    }
    
    return String(result)
}
```

func stringXor(a: String, b: String) -> String {
    let aInt = Int(a, radix: 2)!
    let bInt = Int(b, radix: 2)!
    let xorResult = aInt ^ bInt
    return String(xorResult, radix: 2)
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aInt = Int(a, radix: 2)!
    let bInt = Int(b, radix: 2)!
    let xorResult = aInt ^ bInt
    return String(xorResult, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    let aArray = Array(a.pad(toLength: maxLength, withHeading: "0"))
    let bArray = Array(b.pad(toLength: maxLength, withHeading: "0"))
    
    let xorResult = aArray.indices.map { i in
        let aBit = aArray[i] == "1" ? 1 : 0
        let bBit = bArray[i] == "1" ? 1 : 0
        return String(aBit ^ bBit)
    }
    
    return xorResult.joined()
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    let aArray = Array(a.pad(toLength: maxLength, withHeading: "0"))
    let bArray = Array(b.pad(toLength: maxLength, withHeading: "0"))
    
    let xorResult = aArray.indices.map { i in
        let aBit = aArray[i] == "1" ? 1 : 0
        let bBit = bArray[i] == "1" ? 1 : 0
        return String(aBit ^ bBit)
    }
    
    return xorResult.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" }
    let b = b.map { $0 == "1" }
    let maxLength = max(a.count, b.count)
    var result = [Bool](repeating: false, count: maxLength)
    for i in 0..<maxLength {
        let aIndex = a.count - 1 - i
        let bIndex = b.count - 1 - i
        if aIndex >= 0 && bIndex >= 0 {
            result[i] = a[aIndex] != b[bIndex]
        } else if aIndex >= 0 {
            result[i] = a[aIndex]
        } else if bIndex >= 0 {
            result[i] = b[bIndex]
        }
    }
    return result.map { $0 ? "1" : "0" }.joined()
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" }
    let b = b.map { $0 == "1" }
    let maxLength = max(a.count, b.count)
    var result = [Bool](repeating: false, count: maxLength)
    for i in 0..<maxLength {
        let aIndex = a.count - 1 - i
        let bIndex = b.count - 1 - i
        if aIndex >= 0 && bIndex >= 0 {
            result[i] = a[aIndex] != b[bIndex]
        } else if aIndex >= 0 {
            result[i] = a[aIndex]
        } else if bIndex >= 0 {
            result[i] = b[bIndex]
        }
    }
    return result.map { $0 ? "1" : "0" }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aInt = Int(a, radix: 2)!
    let bInt = Int(b, radix: 2)!
    let xorResult = aInt ^ bInt
    return String(xorResult, radix: 2)
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aInt = Int(a, radix: 2)!
    let bInt = Int(b, radix: 2)!
    let xorResult = aInt ^ bInt
    return String(xorResult, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" }
    let b = b.map { $0 == "1" }
    let maxLength = max(a.count, b.count)
    var result = [Bool](repeating: false, count: maxLength)
    for i in 0..<maxLength {
        let aIndex = a.count - i - 1
        let bIndex = b.count - i - 1
        result[aIndex] = aIndex < a.count && a[aIndex] != b[bIndex]
    }
    return result.map { $0 ? "1" : "0" }.joined()
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" }
    let b = b.map { $0 == "1" }
    let maxLength = max(a.count, b.count)
    var result = [Bool](repeating: false, count: maxLength)
    for i in 0..<maxLength {
        let aIndex = a.count - i - 1
        let bIndex = b.count - i - 1
        result[aIndex] = aIndex < a.count && a[aIndex] != b[bIndex]
    }
    return result.map { $0 ? "1" : "0" }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aInt = Int(a, radix: 2)!
    let bInt = Int(b, radix: 2)!
    let xorResult = aInt ^ bInt
    return String(xorResult, radix: 2)
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aInt = Int(a, radix: 2)!
    let bInt = Int(b, radix: 2)!
    let xorResult = aInt ^ bInt
    return String(xorResult, radix: 2)
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    let aArray = Array(a.pad(toLength: maxLength, withHeading: "0"))
    let bArray = Array(b.pad(toLength: maxLength, withHeading: "0"))
    
    let result = aArray.indices.map { i in
        let aBit = Int(aArray[i].description)
        let bBit = Int(bArray[i].description)
        return String(aBit! ^ bBit!)
    }
    
    return result.joined()
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    let aArray = Array(a.pad(toLength: maxLength, withHeading: "0"))
    let bArray = Array(b.pad(toLength: maxLength, withHeading: "0"))
    
    let result = aArray.indices.map { i in
        let aBit = Int(aArray[i].description)
        let bBit = Int(bArray[i].description)
        return String(aBit! ^ bBit!)
    }
    
    return result.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" }
    let b = b.map { $0 == "1" }
    let maxLength = max(a.count, b.count)
    var result = [Bool]()
    for i in 0..<maxLength {
        let aIndex = a.count - 1 - i
        let bIndex = b.count - 1 - i
        let aBit = aIndex >= 0 ? a[aIndex] : false
        let bBit = bIndex >= 0 ? b[bIndex] : false
        result.append(aBit != bBit)
    }
    return result.map { $0 ? "1" : "0" }.joined()
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" }
    let b = b.map { $0 == "1" }
    let maxLength = max(a.count, b.count)
    var result = [Bool]()
    for i in 0..<maxLength {
        let aIndex = a.count - 1 - i
        let bIndex = b.count - 1 - i
        let aBit = aIndex >= 0 ? a[aIndex] : false
        let bBit = bIndex >= 0 ? b[bIndex] : false
        result.append(aBit != bBit)
    }
    return result.map { $0 ? "1" : "0" }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" }
    let b = b.map { $0 == "1" }
    let maxLength = max(a.count, b.count)
    var result = [Bool]()
    for i in 0..<maxLength {
        let aIndex = a.count - 1 - i
        let bIndex = b.count - 1 - i
        let aBit = aIndex >= 0 ? a[aIndex] : false
        let bBit = bIndex >= 0 ? b[bIndex] : false
        result.append(aBit != bBit)
    }
    return result.map { $0 ? "1" : "0" }.joined()
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let a = a.map { $0 == "1" }
    let b = b.map { $0 == "1" }
    let maxLength = max(a.count, b.count)
    var result = [Bool]()
    for i in 0..<maxLength {
        let aIndex = a.count - 1 - i
        let bIndex = b.count - 1 - i
        let aBit = aIndex >= 0 ? a[aIndex] : false
        let bBit = bIndex >= 0 ? b[bIndex] : false
        result.append(aBit != bBit)
    }
    return result.map { $0 ? "1" : "0" }.joined()
}
```

func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    
    var result = ""
    
    for i in 0..<aArray.count {
        let aBit = Int(aArray[i].description)!
        let bBit = Int(bArray[i].description)!
        
        let xor = aBit ^ bBit
        result += String(xor)
    }
    
    return result
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let aArray = Array(a)
    let bArray = Array(b)
    
    var result = ""
    
    for i in 0..<aArray.count {
        let aBit = Int(aArray[i].description)!
        let bBit = Int(bArray[i].description)!
        
        let xor = aBit ^ bBit
        result += String(xor)
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    }
    if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    }
    if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    var aIndex = a.count - 1
    var bIndex = b.count - 1
    
    var carry = 0
    
    for _ in 0..<maxLength {
        let aBit = aIndex >= 0 ? Int(String(a[aIndex])) : 0
        let bBit = bIndex >= 0 ? Int(String(b[bIndex])) : 0
        
        let xorResult = aBit != bBit ? 1 : (aBit == bBit && carry == 1) ? 1 : 0
        
        carry = aBit != bBit && carry == 1 || aBit == bBit && carry == 1 ? 1 : 0
        
        result.append(String(xorResult))
        
        aIndex -= 1
        bIndex -= 1
    }
    
    if carry == 1 {
        result.append(String(carry))
    }
    
    return result.reversed()
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    var aIndex = a.count - 1
    var bIndex = b.count - 1
    
    var carry = 0
    
    for _ in 0..<maxLength {
        let aBit = aIndex >= 0 ? Int(String(a[aIndex])) : 0
        let bBit = bIndex >= 0 ? Int(String(b[bIndex])) : 0
        
        let xorResult = aBit != bBit ? 1 : (aBit == bBit && carry == 1) ? 1 : 0
        
        carry = aBit != bBit && carry == 1 || aBit == bBit && carry == 1 ? 1 : 0
        
        result.append(String(xorResult))
        
        aIndex -= 1
        bIndex -= 1
    }
    
    if carry == 1 {
        result.append(String(carry))
    }
    
    return result.reversed()
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    let aPadded = String(a.prefix(maxLength))
    let bPadded = String(b.prefix(maxLength))
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = aPadded[i] == "1" ? "1" : "0"
        let bBit = bPadded[i] == "1" ? "1" : "0"
        
        switch (aBit, bBit) {
        case ("1", "1"):
            result.append("1")
        case ("0", "0"):
            result.append("0")
        default:
            result.append("1")
        }
    }
    
    return result
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    let aPadded = String(a.prefix(maxLength))
    let bPadded = String(b.prefix(maxLength))
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = aPadded[i] == "1" ? "1" : "0"
        let bBit = bPadded[i] == "1" ? "1" : "0"
        
        switch (aBit, bBit) {
        case ("1", "1"):
            result.append("1")
        case ("0", "0"):
            result.append("0")
        default:
            result.append("1")
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    } else if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    } else if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = a.index(a.startIndex, offsetBy: i) == "1" ? "1" : "0"
        let bBit = b.index(b.startIndex, offsetBy: i) == "1" ? "1" : "0"
        
        switch (aBit, bBit) {
        case ("1", "1"):
            result.append("1")
        case ("0", "0"):
            result.append("0")
        default:
            result.append("1")
        }
    }
    
    return result
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = a.index(a.startIndex, offsetBy: i) == "1" ? "1" : "0"
        let bBit = b.index(b.startIndex, offsetBy: i) == "1" ? "1" : "0"
        
        switch (aBit, bBit) {
        case ("1", "1"):
            result.append("1")
        case ("0", "0"):
            result.append("0")
        default:
            result.append("1")
        }
    }
    
    return result
}

print(stringXor(a: "010", b: "110")) // "100"
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = a.count > i ? String(a[a.index(a.startIndex, offsetBy: i)]) : ""
        let bBit = b.count > i ? String(b[b.index(b.startIndex, offsetBy: i)]) : ""
        
        let xorResult = aBit == bBit ? "0" : "1"
        result.append(xorResult)
    }
    
    return result
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    
    for i in 0..<maxLength {
        let aBit = a.count > i ? String(a[a.index(a.startIndex, offsetBy: i)]) : ""
        let bBit = b.count > i ? String(b[b.index(b.startIndex, offsetBy: i)]) : ""
        
        let xorResult = aBit == bBit ? "0" : "1"
        result.append(xorResult)
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a.count > i ? String(a[a.index(a.startIndex, offsetBy: i)]) : ""
        let bBit = b.count > i ? String(b[b.index(b.startIndex, offsetBy: i)]) : ""
        
        let xorResult = aBit == bBit ? "0" : "1"
        result.append(xorResult)
    }
    
    return result
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    for i in 0..<maxLength {
        let aBit = a.count > i ? String(a[a.index(a.startIndex, offsetBy: i)]) : ""
        let bBit = b.count > i ? String(b[b.index(b.startIndex, offsetBy: i)]) : ""
        
        let xorResult = aBit == bBit ? "0" : "1"
        result.append(xorResult)
    }
    
    return result
}

print(stringXor(a: "010", b: "110")) // "100"
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    var aIndex = a.count - 1
    var bIndex = b.count - 1
    
    var carry = 0
    for _ in 0..<maxLength {
        let aBit = aIndex >= 0 ? Int(String(a[aIndex])) : 0
        let bBit = bIndex >= 0 ? Int(String(b[bIndex])) : 0
        
        let xorResult = aBit != bBit ? 1 : (aBit == bBit && carry == 1) ? 1 : 0
        carry = aBit != bBit && carry == 1 || aBit == bBit && carry == 1 ? 1 : 0
        
        result.append(String(xorResult))
        
        aIndex -= 1
        bIndex -= 1
    }
    
    if carry == 1 {
        result.append(String(carry))
    }
    
    return result.reversed()
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    
    var result = ""
    var aIndex = a.count - 1
    var bIndex = b.count - 1
    
    var carry = 0
    for _ in 0..<maxLength {
        let aBit = aIndex >= 0 ? Int(String(a[aIndex])) : 0
        let bBit = bIndex >= 0 ? Int(String(b[bIndex])) : 0
        
        let xorResult = aBit != bBit ? 1 : (aBit == bBit && carry == 1) ? 1 : 0
        carry = aBit != bBit && carry == 1 || aBit == bBit && carry == 1 ? 1 : 0
        
        result.append(String(xorResult))
        
        aIndex -= 1
        bIndex -= 1
    }
    
    if carry == 1 {
        result.append(String(carry))
    }
    
    return result.reversed()
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    }
    if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        let aBit = a[String(a[i])] == "1" ? "1" : "0"
        let bBit = b[String(b[i])] == "1" ? "1" : "0"
        result += String(aBit == bBit ? "0" : "1")
    }
    
    return result
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    }
    if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        let aBit = a[String(a[i])] == "1" ? "1" : "0"
        let bBit = b[String(b[i])] == "1" ? "1" : "0"
        result += String(aBit == bBit ? "0" : "1")
    }
    
    return result
}
```

func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros at the beginning
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    }
    if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        let aBit = a[String(a[i])] == "1" ? "1" : "0"
        let bBit = b[String(b[i])] == "1" ? "1" : "0"
        result += String(aBit == bBit ? "0" : "1")
    }
    
    return result
}

You are an AI programming assistant, utilizing the Deepseek Coder model, developed by Deepseek Company, and you only answer questions related to computer science. For politically sensitive questions, security and privacy issues, and other non-computer science questions, you will refuse to answer
### Instruction:
Write code in Swift programming language. Don't include usage examples or tests.
Input are two strings a and b consisting only of 1s and 0s.
Perform binary XOR on these inputs and return result also as a string.
stringXor(a: "010", b: "110") == "100"
### Response:
```swift
func stringXor(a: String, b: String) -> String {
    let maxLength = max(a.count, b.count)
    var result = ""
    
    // Pad shorter string with zeros at the beginning
    if a.count < maxLength {
        a = String(repeating: "0", maxLength - a.count) + a
    }
    if b.count < maxLength {
        b = String(repeating: "0", maxLength - b.count) + b
    }
    
    // Perform XOR operation
    for i in 0..<maxLength {
        let aBit = a[String(a[i])] == "1" ? "1" : "0"
        let bBit = b[String(b[i])] == "1" ? "1" : "0"
        result += String(aBit == bBit ? "0" : "1")
    }
    
    return result
}

print(stringXor(a: "010", b: "110")) // prints "100"
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        let bChar = b[b.index(b.startIndex, offsetBy: index)]
        result += String(char!= bChar? "1" : "0")
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        let bChar = b[b.index(b.startIndex, offsetBy: index)]
        result += String(char!= bChar? "1" : "0")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == b[index] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        if char == b[index] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        result += String(i!= j? "1" : "0")
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        result += String(i!= j? "1" : "0")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[i] == b[i] {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        result += String(i!= j? "1" : "0")
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, j) in zip(a, b) {
        result += String(i!= j? "1" : "0")
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    // TODO: Write your code here
    return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // TODO: Write your code here
    return ""
}
```

func stringXor(a: String, b: String) -> String {
    return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    return ""
}
```

func stringXor(a: String, b: String) -> String {
    // TODO: Implement this function
    return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    // TODO: Implement this function
    return ""
}
```

func stringXor(a: String, b: String) -> String {
    return ""
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    return ""
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        let charB = b.index(b.startIndex, offsetBy: i)
        let charB = b[charB]
        if char == charB {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        let charB = b.index(b.startIndex, offsetBy: i)
        let charB = b[charB]
        if char == charB {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        let charB = b.index(b.startIndex, offsetBy: index)
        if char == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (index, char) in a.enumerated() {
        let charB = b.index(b.startIndex, offsetBy: index)
        if char == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[a.index(a.startIndex, offsetBy: i)]
        let charB = b[b.index(b.startIndex, offsetBy: i)]
        let xor = (charA == "1"? 1 : 0) ^ (charB == "1"? 1 : 0)
        result += String(xor)
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[a.index(a.startIndex, offsetBy: i)]
        let charB = b[b.index(b.startIndex, offsetBy: i)]
        let xor = (charA == "1"? 1 : 0) ^ (charB == "1"? 1 : 0)
        result += String(xor)
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        let char2 = b.index(b.startIndex, offsetBy: i)
        let char2 = b[char2]
        if char == char2 {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for (i, char) in a.enumerated() {
        let char2 = b.index(b.startIndex, offsetBy: i)
        let char2 = b[char2]
        if char == char2 {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[a.index(a.startIndex, offsetBy: i)]
        let charB = b[b.index(b.startIndex, offsetBy: i)]
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[a.index(a.startIndex, offsetBy: i)]
        let charB = b[b.index(b.startIndex, offsetBy: i)]
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        if a[a.index(a.startIndex, offsetBy: i)] == b[b.index(b.startIndex, offsetBy: i)] {
            result += "0"
        } else {
            result += "1"
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[a.index(a.startIndex, offsetBy: i)]
        let charB = b[b.index(b.startIndex, offsetBy: i)]
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}

```swift
// Input are two strings a and b consisting only of 1s and 0s.
// Perform binary XOR on these inputs and return result also as a string.
// stringXor(a: "010", b: "110") == "100"
func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[a.index(a.startIndex, offsetBy: i)]
        let charB = b[b.index(b.startIndex, offsetBy: i)]
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}
```

func stringXor(a: String, b: String) -> String {
    var result = ""
    for i in 0..<a.count {
        let charA = a[a.index(a.startIndex, offsetBy: i)]
        let charB = b[b.index(b.startIndex, offsetBy: i)]
        if charA == charB {
            result.append("0")
        } else {
            result.append("1")
        }
    }
    return result
}