Having trouble with the modulus operator in mASM?

Question

I am trying to implement a pseudo random number generator using modulus in masm. The only problem is I am having trouble with the syntax of assembly.

Here is the code that I wrote in C++:

#include
#include
using namespace std;

int random(int max)
{
int seed = timeGetTime();
return (seed % max);
}

int main()
{
cout << random(4);
return 0;
}

Now, I am having trouble implementing modulus in mASM

To my understanding, the following function should be the following in assembly:

.DATA?
seed DWORD ?

.CODE
start:
call timeGetTime
mov seed, eax
;to my understanding i have to call the MOD function. However, the following does not work:
MOD seed, 4
end start

So my question is, how can I preform the modulus operator in mASM so that I can generate a random number?

Answer 1

Division produces both an integer quotient and remainder, so use DIV for unsigned remainder or IDIV (the usual remainder function implemented by C as the % operator). In either case, EDX and EAX have the 64-bit dividend (numerator) with the high 32 bits in EDX.



So, signed integer division with 32-bit arguments to produce ARGX/ARGY and ARGX%ARGY looks like:



MOV EAX,ARGX ; Get 32-bit signed dividend into EAX

CDQ ; sign extend into EDX

IDIV ARGY ; Now EAX = ARGX/ARGY and EDX = ARGX%ARGY



Unsigned 32-bit division/remainder looks like:



MOV EAX,ARGX ; Get 32-bit unsigned dividend into EAX

SUB EDX,EDX ; zero extend into EDX

IDIV ARGY ; Now EAX = ARGX/ARGY and EDX = ARGX%ARGY

Answer 2

You could always look at the C++ generated code in the debugger and see what it does:



return (seed % max);

00C81680 mov eax,dword ptr [seed]

00C81683 cdq

00C81684 idiv eax,dword ptr [max]

00C81687 mov eax,edx

Answer 3

MOD is not an opcode - it's a MASM operator for assembly-time expressions - you need to understand the difference.



Just use C's % operator - it's going to be a lot easier.

Answer 4

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