Question

Hi
I was interested in what assembly code will gcc generate from this code (this is just dummy code to illustrate my point):

int a = 0;
int foo(void)
{
    int result = a;
    a += 2;
    return result;
}

I was surprized that gcc copies the variable a to the stack and then from the stack to a register so it can return it. When I added register to the declaration of result, it optimized the code not to use the stack but instead to copy the variable directly to a register. I know this doesn’t really make any difference but I was wondering is there any good reason why gcc doesn’t make such optimization implicitly. I hope I made it clear what I am talking about…

Any ideas?

Answer 1

When compiling Debug builds (i.e., with optimizations off), compilers tend to make very straightforward, easily debuggable code. In this case, it might mean keeping all variables in memory / stack, rather than in registers.

Try compiling with full optimizations (-O3) and see if that makes a difference.