Checking if a Bit is Set

To check if bit $k$ is set (equals 1) in number $n$, you AND $n$ with $2^k$. If the result is non-zero, the bit is set.

Why does this work? The number $2^k$ has only one bit set: bit $k$. When you AND with $n$, all other bits become 0. Only bit $k$ survives if it was 1 in $n$.

Example: Check if bit 2 is set in $13 = 1101$. Compute $13 \& (1 << 2) = 13 \& 4 = 1101 \& 0100 = 0100 = 4$. Non-zero, so bit 2 is set.