Why calling printf result in a different function prologue for main? - c

Question

When I compile
#include <stdio.h>
int
main () {
return 0;
}
to x86 assembly the result is plain and expected:
$> cc -m32 -S main.c -o -|sed -r "/\s*\./d"
main:
pushl %ebp
movl %esp, %ebp
movl $0, %eax
popl %ebp
ret
But when studying different disassembled binaries, the function prologue is never that simple. Indeed, changing the C source above into
#include <stdio.h>
int
main () {
printf("Hi");
return 0;
}
the result is
$> cc -m32 -S main.c -o -|sed -r "/\s*\./d"
main:
leal 4(%esp), %ecx
andl $-16, %esp
pushl -4(%ecx)
pushl %ebp
movl %esp, %ebp
pushl %ecx
subl $4, %esp
subl $12, %esp
call printf
addl $16, %esp
movl $0, %eax
movl -4(%ebp), %ecx
leave
leal -4(%ecx), %esp
ret
In particular, I don't get why these instructions
leal 4(%esp), %ecx
andl $-16, %esp
pushl -4(%ecx)
are generated -- specifically why not directly storing %esp into %ecx, instead of into%esp+4?

Answer 1

If main isn't a leaf function, it needs to align the stack for the benefit of any functions it calls. Functions that aren't called main just maintain the stack's alignment.
lea 4(%esp), %ecx # ecx = esp+4
andl $-16, %esp
pushl -4(%ecx) # load from ecx-4 and push that
It's pushing a copy of the return address, so it will be in the right place after aligning the stack. You're right, a different sequence would be more sensible:
mov (%esp), %ecx ; or maybe even pop %ecx
andl $-16, %esp
push %ecx ; push (mem) is slower than push reg
As Youka says in comments, don't expect code from -O0 to be optimized at all. Use -Og for optimizations that don't interfere with debugability. The gcc manual recommends that for compile/debug/edit cycles. -O0 output is harder to read / understand / learn from than optimized code. It's easier to map back to the source, but it's terrible code.