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.

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I wrote this because I felt like it.
This post is my own, and not associated with my employer.
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