Where is the "2+2" in this Assembly code (translated by gcc from C)

Question

I've written this simple C code

int main()
{
    int calc = 2+2;
    return 0;
}

And I want to see how that looks in assembly, so I compiled it using gcc

$ gcc -S -o asm.s test.c

And the result was ~65 lines (Mac OS X 10.8.3) and I only found these to be related:

Where do I look for my 2+2 in this code?

Edit:

One part of the question hasn't been addressed.

If %rbp, %rsp, %eax are variables, what values do they attain in this case?

Answer 1

Almost all of the code you got is just useless stack manipulation. With optimization on (gcc -S -O2 test.c) you will get something like

main:
.LFB0:
    .cfi_startproc
    xorl    %eax, %eax
    ret
    .cfi_endproc
.LFE0:

Ignore every line that starts with a dot or ends with a colon: there are only two assembly instructions:

    xorl %eax, %eax
    ret

and they encode return 0;. (XORing a register with itself sets it to all-bits-zero. Function return values go in register %eax per the x86 ABI.) Everything to do with your int calc = 2+2; has been discarded as unused.

If you changed your code to

int main(void) { return 2+2; }

you would instead get

    movl $4, %eax
    ret

where the 4 comes from the compiler doing the addition itself rather than making the generated program do it (this is called constant folding).

Perhaps more interesting is if you change the code to

int main(int argc, char **argv) { return argc + 2; }

then you get

    leal    2(%rdi), %eax
    ret

which is doing some real work at runtime! In the 64-bit ELF ABI, %rdi holds the first argument to the function, argc in this case. leal 2(%rdi), %eax is x86 assembly language for "%eax = %edi + 2" and it's being done this way mainly because the more familiar add instruction takes only two arguments, so you can't use it to add 2 to %rdi and put the result in %eax all in one instruction. (Ignore the difference between %rdi and %edi for now.)