问题
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?
回答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.)
回答2:
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
来源：https://stackoverflow.com/questions/7703778/how-is-the-modulo-operator-actually-computed