How is the modulo operator (%) actually computed?

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

Answer 1

It uses the idiv assembly instruction:

    int x = a % b;
00000050  cmp         dword ptr [rsp+20h],80000000h 
00000058  jne         0000000000000061 
0000005a  cmp         dword ptr [rsp+24h],0FFFFFFFFh 
0000005f  je          0000000000000070 
00000061  mov         eax,dword ptr [rsp+20h] 
00000065  cdq              
00000066  idiv        eax,dword ptr [rsp+24h] 
0000006a  mov         dword ptr [rsp+2Ch],edx 
0000006e  jmp         0000000000000075 
00000070  call        FFFFFFFFF2620E70 
00000075  mov         eax,dword ptr [rsp+2Ch] 
00000079  mov         dword ptr [rsp+28h],eax 

idiv stores the remainder in a register.
http://pdos.csail.mit.edu/6.828/2007/readings/i386/IDIV.htm

Answer 2

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.)