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.
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.
I wrote this because I felt like it. This post is my own, and not associated with my employer.Jim. Public speaking. Friends. Vidrio.