Clearing a Bit

To clear bit $k$ (make it 0) in number $n$, you AND $n$ with the complement of $2^k$. In code: $n \& \sim(1 << k)$.

Why? The expression $\sim(1 << k)$ creates a number with all bits set to 1 except bit $k$, which is 0. ANDing with this mask turns off bit $k$ while preserving all others.

Example: Clear bit 2 in $13 = 1101$. Compute $13 \& \sim4 = 1101 \& 1011 = 1001 = 9$. Bit 2 is now off.