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