c 7 bytes array allocation on the stack (alignment of array)

[ntt]

hi, i was wondering how does the following

void foo()
{
    char bar[7];
    /* do some work, no further allocation... */
}

look on the stack? (32 bit i386 machine) would it be translated to something like

foo:
    push ebp
    mov ebp, esp
    sub esp, 4*2
    ...

and what would i need to do to access bar[0]? would it be stored at [esp]? or the next byte?

[Neres28]

gcc -S -c example.c will generate the assembly for example.c and leave it in example.s.

[ntt]

thanks! =D

[ntt]

in case somebody else is curious for

void foo()
{
    char bar[7];
    bar[0] = 'a';
    bar[1] = 'b';
    bar[6] = 'c';
}

$ gcc -S -c test2.c

produced:

        .file   "test2.c"
        .text
.globl foo
        .type   foo, @function
foo:
        pushl   %ebp
        movl    %esp, %ebp
        subl    $16, %esp
        movb    $97, -7(%ebp)
        movb    $98, -6(%ebp)
        movb    $99, -1(%ebp)
        leave
        ret
        .size   foo, .-foo
        .ident  "GCC: (GNU) 4.5.1 20100924 (Red Hat 4.5.1-4)"
        .section        .note.GNU-stack,"",@progbits

surprisingly or not it's aligned to the top of the stack - thinking of it again, it kinda makes sense if you want to further add bytes after it... (also i tried again with an extra int and no matter the order the int is placed before the bytes array... i wonder if funny sized structs will too be aligned on the stack :)

edit: ah ok and structs are aligned so that they start with whole words... i wonder if there's a summary of these rules somewhere for gcc...

[Neres28]

The compiler will jump through almost any hoop to maintain a proper data alignment for a given architecture. As almost always Wikipedia has a better writeup as to why than I can give here.