question about gcc assembly on i386

["William D. Colburn (aka Schlake)" <wcolburn at nmt dot edu>]

I have gcc version 2.95.3 20010315 (release) running on a linux ix86 box.

If I compile the following simple program with -S:

-----cut here-----
void hello(void)
{
  printf("hello world\n");
}
-----cut here-----

I get this assembly:

-----cut here-----
	.file	"hello1.c"
	.version	"01.01"
gcc2_compiled.:
.section	.rodata
.LC0:
	.string	"hello world\n"
.text
	.align 4
.globl hello
	.type	 hello,@function
hello:
	pushl %ebp
	movl %esp,%ebp
	subl $8,%esp
	addl $-12,%esp
	pushl $.LC0
	call printf
	addl $16,%esp
.L2:
	movl %ebp,%esp
	popl %ebp
	ret
.Lfe1:
	.size	 hello,.Lfe1-hello
	.ident	"GCC: (GNU) 2.95.3 20010315 (release)"
-----cut here-----	 

There are two lines which make no sense to me:
	subl $8,%esp
	addl $-12,%esp

Why does it add these two lines?  Why are they added as two lines?
Changing the optimization doesn't seem to affect the code generated for
this program at all.

After the printf call it does this:
	addl $16,%esp

So it has recovered the 4 bytes that was pushed onto the stack, plus the
twelve bytes the addl put onto the stack.  That makes it seem like there
needs to be 12 bytes on the stack to make a function call, but I don't
know of any reason that linux/x86s need 12 bytes on the stack.  The call
needs to store an address on the stack, but it pushes it and doesn't
expect the user to provide the space.

So, does anyone have any clue what gcc is doing, and why?

--
William Colburn, "Sysprog" <wcolburn@nmt.edu>
Computer Center, New Mexico Institute of Mining and Technology
http://www.nmt.edu/tcc/     http://www.nmt.edu/~wcolburn