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: 
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.)
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>