Count Bits - Brian Kernighan Trick

A faster approach: repeatedly flip the rightmost 1 bit to 0 until $n$ becomes 0. The number of flips equals the number of 1 bits.

The operation $n \& (n - 1)$ flips the rightmost 1 bit to 0. Why? Subtracting 1 flips all trailing bits up to and including the rightmost 1. ANDing with $n$ zeros out that rightmost 1. For example, $1100 \& 1011 = 1000$.

Loop: while $n > 0$, do $n := n \& (n - 1)$ and increment a counter. When $n$ reaches 0, all 1s have been flipped. This runs in $O(k)$ where $k$ is the number of 1 bits, not $O(32)$.