C – Understanding Base Pointer and Stack Pointers in GCC Output

assemblyc++gccgnu-assembler

I have the following C program:

int main()
{
    int c[10] = {0, 0, 0, 0, 0, 0, 0, 0, 1, 2};
    return c[0];
}

and when compiled using the -S directive with gcc I get the following assembly:

    .file   "array.c"
    .text
.globl main
    .type   main, @function
main:
.LFB0:
    .cfi_startproc
    pushq   %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset 6, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register 6
    movl    $0, -48(%rbp)
    movl    $0, -44(%rbp)
    movl    $0, -40(%rbp)
    movl    $0, -36(%rbp)
    movl    $0, -32(%rbp)
    movl    $0, -28(%rbp)
    movl    $0, -24(%rbp)
    movl    $0, -20(%rbp)
    movl    $1, -16(%rbp)
    movl    $2, -12(%rbp)
    movl    -48(%rbp), %eax
    leave
    .cfi_def_cfa 7, 8
    ret
    .cfi_endproc
.LFE0:
    .size   main, .-main
    .ident  "GCC: (GNU) 4.4.5 20110214 (Red Hat 4.4.5-6)"
    .section        .note.GNU-stack,"",@progbits

What I do not understand is why are the earlier array elements further from the bp? It almost seems like the elements on the array are being placed in opposite order.

Also why does gcc not use push instead of movl, to push the array elements onto the stack?

DIFFERENT VIEW

Moving the array to global namespace as a static variable to the module I get:

    .file   "array.c"
    .data
    .align 32
    .type   c, @object
    .size   c, 40
c:
    .long   0
    .long   0
    .long   0
    .long   0
    .long   0
    .long   0
    .long   0
    .long   0
    .long   1
    .long   2
    .text
.globl main
    .type   main, @function
main:
.LFB0:
    .cfi_startproc
    pushq   %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset 6, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register 6
    movl    c(%rip), %eax
    leave
    .cfi_def_cfa 7, 8
    ret
    .cfi_endproc
.LFE0:
    .size   main, .-main
    .ident  "GCC: (GNU) 4.4.5 20110214 (Red Hat 4.4.5-6)"
    .section    .note.GNU-stack,"",@progbits

Using the following C program:

static int c[10] = {0, 0, 0, 0, 0, 0, 0, 0, 1, 2};

int main() 
{
    return c[0];
}

This doesn't give more insight to the stack. But it is intersting to see the differement output of assembly using slightly different semantics.

Best Answer

First of all, the x86 stack grows downwards. By convention, rbp stores the original value of rsp. Therefore, the function's arguments reside at positive offsets relative to rbp, and its automatic variables reside at negative offsets. The first element of an automatic array has a lower address than all other elements, and thus is the furthest away from rbp.

Here is a handy diagram that appears on this page:

stack layout

I see no reason why the compiler couldn't use a series of push instructions to initialize your array. Whether this would be a good idea, I am not sure.

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