Or literally just look at its binary representation. If the least significant digit is a "1", it's odd, if "0", it's even. Or you can divide by 2 and check for a remainder.
Your method is just spending time grinding away CPU cycles for no reason.
Or literally just look at its binary representation. If the least significant digit is a "1", it's odd, if "0", it's even. Or you can divide by 2 and check for a remainder.
Your method is just spending time grinding away CPU cycles for no reason.