What's the sign of C's remainder operator "%"?

July 12, 2020

TL;DR In C99 or later, the sign of a%b is the same as a; earlier than C99 it is implementation-defined.

The standard says the sign of a%b is such that (a/b)*b + a%b == a. In other words, a%b == a - (a/b)*b. Assuming C99, a/b always rounds towards zero; so a/b is no larger in magnitude than its correct value. This means |(a/b)*b| <= a (1).

(a/b)*b has the same sign as a, because the sign of b cancels out. With this and (1), we have that a - (a/b)*b has the same sign as a.

Therefore, a%b has the same sign as a. It is the difference between a and the nearest multiple of b to a in the direction of 0.

Note that prior to C99, the rounding direction of a/b was implementation-defined. This means it may be true that |(a/b)*b| > a, making a - (a/b)*b and hence a%b the opposite sign to a. Hence, prior to C99, the sign of a%b is implementation-defined.

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