How can I do modulo with a bitmask in C?

Are you doing i % n, where n is a power of two? There’s a neat alternative way to do that: i & (n-1). See how it works by choosing n == 8, so i % 8 is the same as i & 7. 7 is 0b111, so i & 0b111 removes all but the three least significant bits. For instance, 14 % 8 == 6. But 14 & 7 == 0b1110 & 0b0111 == 0b0110 == 0b110 = 6.

For example, you might use this when implementing a ring buffer with power-of-two length.

Get updates on Twitter

More by Jim

Tagged . All content copyright James Fisher 2016. This post is not associated with my employer. Found an error? Edit this page.