Integer evenness testing (% 2 vs & 1)

Question

Do these two statements compile equivalently: n % 2 == 0 and n & 1 == 0 ?

if not, is one more efficient?

Answer 1

No, they do not always give the same result. The C standard allows for ones' complement implementations, in which case they will give a different result for negative n.

Answer 2

A similar question was asked yesterday.

i % 2 and i & 1 are not the same:

• If your integer is not unsigned, depending on its representation, n & 1 does not necessarily indicate whether your integer is even or odd.
• i % 2 can be negative.

For unsigned integers, a good compiler should reasonably produce equally efficient code.