What's the sign of C's remainder operator "%"?
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%b has the same sign as
a. It is the difference between
a and the nearest multiple of
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.