[ale] linux byte alignment

[Benjamin Dixon]

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?

Ben


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