Adding doubles in x86_64 assembly problems

Question

Hello I am trying to learn assembly and learn how to work with floating point numbers in x86_64. From what I understand arguments are passed in xmm0, xmm1, xmm2, and so on, and the result is returned in xmm0. So I am trying to make a simple assembly function that adds to double together. Here is the function

.text

.global floatadd
.type floatadd,@function

floatadd:
    addsd %xmm1,%xmm0
    ret

And here is the C code I am using as well.

#include<stdio.h>

int main(){
    double a = 1.5;
    double b = 1.2;
    double c = floatadd(a,b);
    printf("result = %f\n",c);
}

I have been trying to following what is happening in gdb. When I set a breakpoint in my function I can see xmm0 has 1.5 and xmm1 has 1.2 and when they are added together they 2.7. In gdb print $xmm0 gives v2_double = {2.7000000000000002, 0} However when my function returns from main and calls

cvtsi2sd %eax,%xmm0

Print $xmm0 becomes v2_double = {2, 0}. I am not sure why gcc calls that or why it is uses the 32bit register instead of the 64bit register. I have tried using the modifier %lf, and %f and both of them do the same thing.

What is happening?

Where's the prototype for floatadd? If you're not specifying one, your compiler is treating floatadd as returning int, which would explain the use of that instruction. — Peter Huene

Chris Dodd Chris Dodd · Accepted Answer · 2013-10-23T00:55:15

The problem is that you failed to declare floatadd before calling it. So the compiler assumes it returns an int in %eax and converts that int to a double. Add the declaration:

double floatadd(double, double);

before main.

Using -Wall or whatever equivalent your compiler uses to enable warnings would probably have told you about this problem...

Adding doubles in x86_64 assembly problems

1 Answers