Modulo of negative numbers [duplicate]

Answer 1

The solution with two % operations works, but this is somewhat faster in most languages on most hardware (there are exceptions, however):

int wrapIndex(int i, int i_max) {
    i = i%i_max;
    return i<0 ? i+i_max : i;
}

Answer 2

Many users gave good answers, just beware negative numbers, since different languages may behave differently. By example this C snippet writes "-1"

int main ()
{
    printf("%d\n", (-4) % 3);
}

While in python we have a different output value

Python 2.6.4 (r264:75706, Dec  7 2009, 18:43:55) 
[GCC 4.4.1] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> (-4) % 3
2

EDIT: Actually I don't think you'll have negative indexes! However it's good to know that.

Answer 3

This solution is branchless, but performs % twice:

function wrapIndex(i, i_max) {
   return ((i % i_max) + i_max) % i_max;
}

It should be said the C#/Java behavior of % is assumed, i.e. the result has the same sign as the dividend. Some languages define the remainder calculation to take the sign of the divisor instead (e.g. mod in Clojure). Some languages have both variants (mod/rem pair in Common Lisp, Haskell, etc). Algol-68 has %x which always returns a non-negative number. C++ left it up to implementation until C++11, now the sign of the remainder is (almost) fully specified according to the dividend sign.

See also

• Wikipedia/Modulo operation

Answer 4

Nicer is a matter of taste, but How about

var x = (i_max + i % i_max) % i_max;

Answer 5

You could do this:

function wrapIndex(i, i_max) {
    if (i < 0) i = (i % i_max) + i_max;
    return i % i_max;
}