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(n & (1 << k)) ? (n ^ (1 << k)) : n

(n & (1 << k)) ? (n ^ (1 << k)): n

(~(1 << k)) & n

n & (!(1 << k))

Let's keep this example at 8 bits for simplicity.

Let's start with our sample input: input = 10101010
Create another number i which will start at 1: i = 00000001
Shift it k times. (Let's say k = 3): i = 00001000
Negate it with XOR by performing i ^ Int.max: i = i ^ 11111111 = 11110111
Return i & number: 10100010

Swift 2.2

func flip(k: Int, number:Int) -> String {
    var operand = 1
    operand = operand << k
    operand = operand ^ Int.max
    return String(operand & number, radix: 2)
    
}

flip(0, number: 0b101) // "100"
flip(1, number: 0b101) // "101"
flip(2, number: 0b101) // "1"

(n & (1<<k)) ? (n - (1 << k)) : n

(n & (1 << k)) ? (n - (1 << k)): n

isnt it just:

$n ^ (1 << $k) ?