c - gcc 版本 4.6.3 上的 c 程序汇编版本

Question

我用c语言编写了一个小型数组程序

#include <stdio.h>
int main()
{
    int arr[]={1,2,3,4,5};//int size is 4, elements 5 so size of array = 4*5 = 20
    printf("%d\n", sizeof(arr));
    return 0;
}

我用

编译了这个

gcc -O2 -fverbose-asm -S -c arr_n_pointer_confusion.c 

我明白了，

    .section    .rodata.str1.1,"aMS",@progbits,1
.LC0:
    .string "%d\n"
    .section    .text.startup,"ax",@progbits
    .p2align 4,,15
    .globl  main
    .type   main, @function
main:
.LFB22:
    .cfi_startproc
    pushl   %ebp    #
    .cfi_def_cfa_offset 8
    .cfi_offset 5, -8
    movl    %esp, %ebp  #,
    .cfi_def_cfa_register 5
    andl    $-16, %esp  #,
    subl    $16, %esp   #,
    movl    $20, 8(%esp)    #,
    movl    $.LC0, 4(%esp)  #,
    movl    $1, (%esp)  #,
    call    __printf_chk    #
    xorl    %eax, %eax  #
    leave
    .cfi_restore 5
    .cfi_def_cfa 4, 4
    ret
    .cfi_endproc
.LFE22:
    .size   main, .-main
    .ident  "GCC: (Ubuntu/Linaro 4.6.3-1ubuntu5) 4.6.3"
    .section    .note.GNU-stack,"",@progbits

任何人都可以将 C 语言中的汇编步骤联系起来吗？我之所以尝试这样做是为了了解一点汇编代码，以便我能够理解从指针到数组的区别。

Answer 1

基本上，因为编译器在编译时就知道数组的内容，所以它可以删除数组，只需将 sizeof(array) 替换为 20，而不必在实际初始化数组运行时。

.cfi_startproc
pushl   %ebp    # Save the value of ebp (points to the base of the stack) in the stack
.cfi_def_cfa_offset 8
.cfi_offset 5, -8
movl    %esp, %ebp  #, Set the value of the base of the stack to the top of it
.cfi_def_cfa_register 5
andl    $-16, %esp  # Align the stack with a 16-byte boundary, used for SIMD instructions
subl    $16, %esp   # Subtract 16 from the value of the stack pointer (reserving 16 bytes of space for the stack)
movl    $20, 8(%esp)    # Set the memory 8 bytes above the stack as '20'
movl    $.LC0, 4(%esp)  # Move the string "%d\n" 4 bytes above the stack
movl    $1, (%esp)  # Set the flag for __printf_chk() to 1, enabling stack overflow checks
call    __printf_chk    # Print our string
xorl    %eax, %eax # Zero out the eax register (i.e. store the return code in eax, which is 0)

。因此传递给 __printf_chk 的参数是 (1, "%d\n", 20)。