32b x86汇编scanf用法

所以,我试图scanf在32位ATT程序集中使用该函数并继续得到分段错误,尽管使用的代码Computer Systems: A Programmer's Perspective与我自己的简单C输入程序生成的汇编代码几乎相同.我不知道我做错了什么,并希望得到一些帮助.

我的测试汇编代码(segfaults):

    .data
    .align  4
fmt:    .string "%d"
str:    .string "Input a number: "
    .text

    .global main
    .type   main, @function
main:
    pushl   %ebp
    movl    %esp, %ebp

    subl    $40, %esp

    movl    $str, (%esp)
    call    printf

    leal    36(%esp), %eax
    movl    %eax, 4(%esp)
    movl    $fmt, (%esp)
    call    scanf

    pushl   -4(%ebp)
    call    printf

    movl    %ebp, %esp
    popl    %ebp
    ret

C代码和它的汇编:

C:

#include 

int main()
{
    int i, j;
    printf("%s\n","Enter 2 numbers:");
    scanf("%d %d",&i,&j);
    printf("i = %d and j = %d\n",i,j);
    return 0;
}

部件:

    .file   "scan.c"
    .section    .rodata
.LC0:
    .string "Enter 2 numbers:"
.LC1:
    .string "%d %d"
.LC2:
    .string "i = %d and j = %d\n"
    .text
    .globl  main
    .type   main, @function
main:
.LFB0:
    .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    $32, %esp
    movl    $.LC0, (%esp)
    call    puts
    leal    28(%esp), %eax
    movl    %eax, 8(%esp)
    leal    24(%esp), %eax
    movl    %eax, 4(%esp)
    movl    $.LC1, (%esp)
    call    __isoc99_scanf
    movl    28(%esp), %edx
    movl    24(%esp), %eax
    movl    %edx, 8(%esp)
    movl    %eax, 4(%esp)
    movl    $.LC2, (%esp)
    call    printf
    movl    $0, %eax
    leave
    .cfi_restore 5
    .cfi_def_cfa 4, 4
    ret
    .cfi_endproc
.LFE0:
    .size   main, .-main
    .ident  "GCC: (Ubuntu 4.8.4-2ubuntu1~14.04) 4.8.4"
    .section    .note.GNU-stack,"",@progbits

书中的例子(在剪裁的截图中):

你只是忘记了格式字符串printf.你有效地做printf(i)而不是printf("%d", i).因此改变:

pushl   -4(%ebp)
call    printf

至:

pushl   -4(%ebp)
pushl   $fmt
call    printf

PS:学会使用调试器.