How to write modulus (% in C++) in assembly language

Question

I do not know what instruction I have to use in order to translate %2

#include <iostream>
using namespace std;

int main () {
     int number;
     cin >> number;
     if (number % 2 == 0) {    // I cannot translate this part.
       cout << "Even\n";
    }
    else {
       cout << "Odd\n";
    }
    return 0;
 }

compile your program and get assembler output from compiler. Job done. — Tony The Lion, Nov 22 '13 at 16:26
Let the compiler help you :) Translate to assembly using your favorite C++ compiler, and check the output. — Sergey Kalinichenko, Nov 22 '13 at 16:26
Your question is puzzling: do you wish to convert your program in C++ to assembly, _or_ do you want to know what's the equivalent code in assembly for the remainder operation? — jrd1, Nov 22 '13 at 16:26
And hey, an [answer](http://stackoverflow.com/questions/8021772/assembly-language-how-to-do-modulo)! And [another](http://stackoverflow.com/questions/4361979/calculating-modulo-in-assembly)! — crashmstr, Nov 22 '13 at 16:27
What is your target platform? Different processors use different assembly language to perform the operation. This is twice you haven't specified. — Thomas Matthews, Nov 22 '13 at 16:30
Most C compilers have an option to generate assembly listing while compiling. On MSVC, it's /FA. — Seva Alekseyev, Nov 22 '13 at 16:30
Modulo is nothing. If I were you I wound rather panic about hand-translating std::cout::operator<<. — Alex Watson, Nov 22 '13 at 16:34
@Keith Which you need to synthesize yourself, cout being a template. — Alex Watson, Nov 22 '13 at 17:02

Jerry Coffin · Answer 1 · 2013-11-22T16:40:55.733

In a typical assembly language, the integer divide instruction will also give you the remainder. In the case of remainder when dividing by 2, it's a lot easier to translate to a bit-wise AND with bit 0 though. E.g., on x86:

    mov eax, number

    ; test sets the flags based on a bitwise and, but discards the result
    test eax, 1

    mov esi, offset even_string
    jz print_it
    mov esi, offset odd_string
print_it:
    ; print string whose base address is in esi

If you need to check for divisibility by some arbitrary number (instead of just 2), the division instruction will usually produce both the quotient and the remainder. Again, using x86 as a demonstration:

idiv ebx  ; divisor in edx:eax
; quotient in eax, remainder in edx

score 2 · Accepted Answer · answered Nov 22 '13 at 16:28

Modulo is typically produced by the division instruction. However, taking modulo 2, 4, 8, 16, and so on is a special case: because the numbers are stored in the binary form, these checks can be done with an AND instruction.

The fastest way of checking divisibility by 2 is looking at the least significant bit of a value. Use AND instruction with the value of 1, i.e. number & 1. Its result is always the same as number % 2. Most modern compilers (and some archaic ones) will do this as an easy optimization.

For other powers of two, AND with that value minus one, i.e. for x % 4 use x & 3, for x % 8 use x & 7, and so on.

score 0 · Answer 3 · answered Nov 22 '13 at 16:35

In assembly languages that don't have a bit test instruction, you will have to (Boolean) AND the value with 1:

    LODI, R0  #5    ; Load register 0 with the number 5.
    ANDI, R0  #1    ; Boolean AND the register with one (1).
    JUMZ      Value_is_even; jump if the value is zero.
;
;   Go here if the value is odd.

How to write modulus (% in C++) in assembly language

3 Answers3