Linux asm code

I have compiled a useless C module with Gcc under Linux just to see
what compiled C code looks like. The C code is this:

print1(char * str)
{
printf(str);
}

The output asm code is as follows (generated by objdump)

00000013 <print1>:
13: 55 push %ebp
14: 89 e5 mov %esp,%ebp
16: 83 ec 08 sub $0x8,%esp
19: 8b 45 08 mov 0x8(%ebp),%eax
1c: 89 04 24 mov %eax,(%esp)
1f: e8 fc ff ff ff call 20 <print1+0xd>
24: c9 leave
25: c3 ret

The code is relatively simple, but I don't understand the third
instruction (sub 0x8, esp). obviously it allocates 8 bytes to the
stack, but why does it allocate this much? It only uses 4 bytes of it
(to pass the str parameter to printf), and leaves the "top" four
unused. Why is this? Or am I confusing something??

My mental image of the stack just before the call to printf is as
follows. Is this accurate?

|Param to print1 |
|--------------------|
|Addr of caller |
|--------------------|
|Caller EBP |
|--------------------|
|UNUSED DWORD |
|--------------------|
|Param to printf |
|--------------------|

cheers
MQ

.

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