Question

How to check if an integer contains an even or odd number of bits set in Swift

Answer 1

import Foundation

let num = 42 // 00101010 in binary

// Convert number to binary string and count '1's
let binaryStr = String(num, radix: 2)
let bitCount = binaryStr.filter { $0 == "1" }.count
let result = bitCount % 2

print("0 = even number of bits set")
print("1 = odd number of bits set")
print("result: \(result)")



/*
run:

0 = even number of bits set
1 = odd number of bits set
result: 1

*/

 

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