[ale] linux byte alignment

[stephen]

On Mon, Jul 29, 2002 at 03:26:09PM -0400, Benjamin Dixon wrote:
> 
> Hi all,
> 
> I'm trying to pry into linux byte alignment issues and assembly and I ran
> across something I haven't figured out. My understanding is that alignment
> is at one word (4 bytes) so I have the following function:
> 
> int main()
> {
>    char x[10];
>    char y[5];
> }
> 
> By my calculation, if the memory has to be alignment, x[10] will take up
> 12 bytes (ceiling of 2.5 words = 3, 3x4-bytes = 12). And likewise, the
> y[5] will take up 8 bytes. So there's 20 bytes of excess memory laying
> around? But when I run the program through gcc with the -S option, I get
> the following:
> 
> ...
> main:
>         pushl %ebp
>         movl %esp,%ebp
>         subl $40,%esp
> .L2:
>         movl %ebp,%esp
>         popl %ebp
>         ret
> .Lfe1:
> ...
> 
> The question is, what's that 40? If I use different numbers for the array
> sizes, I get a different number there, always divisible by 4 but always
> greater than the number I expect. Anyone know why?

I'm not sure why the compiler produces this, but if you disassemble the
the binary then you should see the expected 20 bytes for x and y plus room
for the frame pointer and return address.  This is a snippet of what I got
from 'objdump -d':

[snip]
080483a4 <main>:
 80483a4:	55                   	push   %ebp
 80483a5:	89 e5                	mov    %esp,%ebp
 80483a7:	83 ec 28             	sub    $0x28,%esp
 80483aa:	c9                   	leave  
 80483ab:	c3                   	ret    
 80483ac:	90                   	nop    
 80483ad:	90                   	nop    
 80483ae:	90                   	nop    
 80483af:	90                   	nop    
[/snip]

stephen

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