i386 assembly question: why do I need to meddle with the stack pointer?

Question

I decided it would be fun to learn x86 assembly during the summer break. So I started with a very simple hello world program, borrowing on free examples gcc -S could give me. I ended up with this:

HELLO:
    .ascii "Hello, world!\12\0"
    .text

.globl _main
_main:
    pushl   %ebp        # 1. puts the base stack address on the stack
    movl    %esp, %ebp  # 2. puts the base stack address in the stack address register
    subl    $20, %esp   # 3. ???
    pushl   $HELLO      # 4. push HELLO's address on the stack
    call    _puts       # 5. call puts
    xorl    %eax, %eax  # 6. zero %eax, probably not necessary since we didn't do anything with it
    leave               # 7. clean up
    ret                 # 8. return
                        # PROFIT!

It compiles and even works! And I think I understand most of it.

Though, magic happens at step 3. Would I remove this line, my program would die between the call to puts and the xor from a misaligned stack error. And would I change $20 to another value, it'd crash too. So I came to the conclusion that this value is very important.

Problem is, I don't know what it does and why it's needed.

Can anyone explain me? (I'm on Mac OS, would it ever matter.)

Answer 1

On x86 OSX, the stack needs to be 16 byte aligned for function calls, see ABI doc here. So, the explanation is

push stack pointer (#1)         -4
strange increment (#3)         -20
push argument (#4)              -4
call pushes return address (#5) -4
total                          -32

To check, change line #3 from $20 to $4, which also works.

Also, Ignacio Vazquez-Abrams points out, #6 is not optional. Registers contain remnants of previous calculations so it has to explicitly be zeroed.

I recently learned (still learning) assembly, too. To save you the shock, 64bit calling conventions are MUCH different (parameters passed on the register). Found this very helpful for 64bit assembly.