Isolating the Rightmost Set Bit

Sometimes you need to isolate the rightmost set bit, turning off all others. The formula is $n \& (-n)$.

Why does this work? In two's complement representation, $-n$ is computed as $\sim n + 1$. When you add 1 to $\sim n$, it flips all the trailing 0s in $n$ to 1s and the rightmost 1 to 0, then carries over. ANDing $n$ with this isolates just that rightmost 1 bit.

Example: $n = 12 = 1100$. Then $-n = -12 = ...0100$ (in two's complement, focusing on the low bits). $n \& (-n) = 1100 \& 0100 = 0100 = 4$. Only the rightmost set bit remains.