 Signed power-of-two division by biasing

For signed two's complement integers, a conditional positive bias converts the rounding behaviour of an arithmetic right shift into C integer division. It can be formed without a conditional branch by turning the sign bit into either zero or a low-$k$-bit bit mask.

Proof. Nonnegative $a$ needs no adjustment. If $a<0$, write $-a=2^{k}q+r$, $0\le r<2^k$. Then $a+(2^k-1)=-2^{k}q+(2^k-1-r)$, whose quotient rounded down is $-q$, exactly the quotient truncated towards zero. The bias fits safely for $1\le k<n$ in an $n$-bit signed word: it is added only to negative inputs, and the largest possible adjusted input is $2^k-2$. The corresponding remainder is $((a+b)\mathrm{mod}{2}^k)-b$, where $b=(2^k-1)1_{a<0}$.