gcc - .comm 是什么意思？

Question

我刚刚翻译了这个程序，

#include <stdio.h>

int dam[1000][1000];


int main (int argc, const char * argv[]) {
    // insert code here...
    printf("Hello, World!\n");
    return 0;
}

使用gcc生产进行组装，

    .cstring
LC0:
    .ascii "Hello, World!\0"
    .text
.globl _main
_main:
    pushl   %ebp
    movl    %esp, %ebp
    pushl   %ebx
    subl    $20, %esp
    call    L3
"L00000000001$pb":
L3:
    popl    %ebx
    leal    LC0-"L00000000001$pb"(%ebx), %eax
    movl    %eax, (%esp)
    call    L_puts$stub
    movl    $0, %eax
    addl    $20, %esp
    popl    %ebx
    leave
    ret
.comm _dam,1000000,5
    .section __IMPORT,__jump_table,symbol_stubs,self_modifying_code+pure_instructions,5
L_puts$stub:
    .indirect_symbol _puts
    hlt ; hlt ; hlt ; hlt ; hlt
    .subsections_via_symbols

.comm 是什么意思？ dam 使用堆空间、堆栈空间还是数据空间？

Answer 1

来自 as手册:

..comm declares a common symbol named symbol. When linking, a common symbol in one object file may be merged with a defined or common symbol of the same name in another object file. If ld does not see a definition for the symbol--just one or more common symbols--then it will allocate length bytes of uninitialized memory. length must be an absolute expression. If ld sees multiple common symbols with the same name, and they do not all have the same size, it will allocate space using the largest size.

When using ELF, the .comm directive takes an optional third argument. This is the desired alignment of the symbol, specified as a byte boundary (for example, an alignment of 16 means that the least significant 4 bits of the address should be zero). The alignment must be an absolute expression, and it must be a power of two. If ld allocates uninitialized memory for the common symbol, it will use the alignment when placing the symbol. If no alignment is specified, as will set the alignment to the largest power of two less than or equal to the size of the symbol, up to a maximum of 16.