Why does gcc use movl instead of push to pass function args?

Question

pay attention to this code :

#include <stdio.h>
void a(int a, int b, int c)
{
    char buffer1[5];
    char buffer2[10];
}

int main()
{
    a(1,2,3); 
}

after that :

gcc -S a.c

that command shows our source code in assembly.

now we can see in the main function, we never use "push" command to push the arguments of the a function into the stack. and it used "movel" instead of that

main:
 pushl %ebp
 movl %esp, %ebp
 andl $-16, %esp
 subl $16, %esp
 movl $3, 8(%esp)
 movl $2, 4(%esp)
 movl $1, (%esp)
 call a
 leave

why does it happen? what's difference between them?

Answer 1

Here is what the gcc manual has to say about it:

-mpush-args
-mno-push-args
    Use PUSH operations to store outgoing parameters. This method is shorter and usually
    equally fast as method using SUB/MOV operations and is enabled by default. 
    In some cases disabling it may improve performance because of improved scheduling
    and reduced dependencies.

 -maccumulate-outgoing-args
    If enabled, the maximum amount of space required for outgoing arguments will be
    computed in the function prologue. This is faster on most modern CPUs because of
    reduced dependencies, improved scheduling and reduced stack usage when preferred
    stack boundary is not equal to 2. The drawback is a notable increase in code size.
    This switch implies -mno-push-args. 

Apparently -maccumulate-outgoing-args is enabled by default, overriding -mpush-args. Explicitly compiling with -mno-accumulate-outgoing-args does revert to the PUSH method, here.

---

2019 update: modern CPUs have had efficient push/pop since about Pentium M.
-mno-accumulate-outgoing-args (and using push) eventually became the default for -mtune=generic in Jan 2014.