Query about working out whether number is a power of 2

Question

Using the classic code snippet:

if (x & (x-1)) == 0

If the answer is 1, then it is false and not a power of 2. However, working on 5 (not a power of 2) a

Answer 1

((n & (n-1)) == 0)

It checks whether the value of “n” is a power of 2.

Example:

 if n = 8, the bit representation is 1000
 n & (n-1) = (1000) & ( 0111) = (0000)
 So it return zero only if its value is in power of 2.
 The only exception to this is ‘0’.
 0 & (0-1) = 0 but ‘0’ is not the power of two. 

Why does this make sense?

Imagine what happens when you subtract 1 from a string of bits. You read from left to right, turning each 0 to a 1 until you hit a 1, at which point that bit is flipped:

1000100100 -> (subtract 1) -> 1000100011

Thus, every bit, up through the first 1, is flipped. If there’s exactly one 1 in the number, then every bit (other than the leading zeros) will be flipped. Thus, n & (n-1) == 0 if there’s exactly one 1. If there’s exactly one 1, then it must be a power of two.