How to check if a number is a power of 2

Question

Today I needed a simple algorithm for checking if a number is a power of 2.

The algorithm needs to be:

1. Simple
2. Correct for any ulong

Answer 1

return ((x != 0) && !(x & (x - 1)));

If x is a power of two, its lone 1 bit is in position n. This means x – 1 has a 0 in position n. To see why, recall how a binary subtraction works. When subtracting 1 from x, the borrow propagates all the way to position n; bit n becomes 0 and all lower bits become 1. Now, since x has no 1 bits in common with x – 1, x & (x – 1) is 0, and !(x & (x – 1)) is true.