Why does gcc push %rbx at the beginning of main?

Question

The latest version of gcc is producing assembly that doesn\'t make sense to me. I compiled the code using no optimization; but, some parts of this code don\'t make sense, even

Answer 1

GCC dictates how the stack is used. Contract between caller and callee on x86:

    * after call instruction:
          o %eip points at first instruction of function
          o %esp+4 points at first argument
          o %esp points at return address 
    * after ret instruction:
          o %eip contains return address
          o %esp points at arguments pushed by caller
          o called function may have trashed arguments
          o %eax contains return value (or trash if function is void)
          o %ecx, %edx may be trashed
          o %ebp, %ebx, %esi, %edi must contain contents from time of call 
    * Terminology:
          o %eax, %ecx, %edx are "caller save" registers
          o %ebp, %ebx, %esi, %edi are "callee save" registers

The main function is like any other function in this context. gcc decided to use ebx for intermediate calculations, so it preserves its value.

Answer 2

By default gcc compiles with optimization disabled, which is the case here, apparently.

You need to enable it with one of the optimization switches (e.g. -O2 or -O3).

Then you will not see redundant and seemingly meaningless things.

As for rbx, it has to be preserved because that's what the calling conventions require. Your function modifies it (movl -32(%rbp), %ebx), so it has to be saved and restored explicitly.