Question

Recently I’ve been confused about the modulo operator, %.

It’s known that a % b == a-a/b*b when we have integers a and b where a > b, and we can do this calculation by hand if a and b are small enough.

However, when it comes to the way a processor computes it, does the processor use the same method as previously mentioned, a-a/b*b? Maybe just by translating the division into subtraction or addition, or is there some shifting involved perhaps?

Answer 1

Except for powers of 2, where the modulo operator can (and in most optimizing compilers is) be turned into a simple bitwise operation, I’m afraid the only way to do it is the hard way. Explanation is http://en.wikipedia.org/wiki/Modulo_operation

In another answer, @Henk Holterman points out that some CPUs do it in the microcode, leaving the remainder in a register while doing an integer divide, which means the modulo instruction can be reduced to an integer divide and return the remainder. (I’m adding that information here because this answer has already been accepted.)