从汇编程序获取命令行参数

时间:2013-08-26 16:06:48

标签: linux assembly x86 gas

阅读“专业汇编语言书”;它似乎提供了一个错误的代码来读取命令行参数。我纠正了一下,现在它从segfaulting到读取参数计数然后是segfaulting。

这是完整的代码:

.data
    output1:
        .asciz "There are %d params:\n"
    output2:
        .asciz "%s\n"

.text
.globl main
main:
    movl 4(%esp), %ecx  /* Get argument count.  */
    pushl %ecx
    pushl $output1
    call printf
    addl $4, %esp       /* remove output1  */

    /* ECX was corrupted by the printf call,
       pop it off the stack so that we get it's original
       value.  */
    popl %ecx

    /* We don't want to corrupt the stack pointer
       as we move ebp to point to the next command-line
       argument.  */
    movl %esp, %ebp
    /* Remove argument count from EBP.  */
    addl $4, %ebp

pr_arg:
    pushl (%ebp)
    pushl $output2
    call printf
    addl $8, %esp       /* remove output2 and current argument.  */
    addl $4, %ebp       /* Jump to next argument.  */
    loop pr_arg

    /* Done.  */
    pushl $0
    call exit

书中的代码:

.section .data
output1:
    .asciz “There are %d parameters:\n”
output2:
    .asciz “%s\n”
.section .text
.globl _start
_start:
    movl (%esp), %ecx
    pushl %ecx
    pushl $output1
    call printf
    addl $4, %esp
    popl %ecx
    movl %esp, %ebp
    addl $4, %ebp
loop1:
    pushl %ecx
    pushl (%ebp)
    pushl $output2
    call printf
    addl $8, %esp
    popl %ecx
    addl $4, %ebp
    loop loop1
    pushl $0
    call exit

用GCC(gcc cmd.S)编译,也许这就是问题所在? __libc_start_main以某种方式修改堆栈?不太确定......

更糟糕的是,尝试调试它以查看堆栈但GDB似乎抛出了许多与printf相关的东西(其中一个是printf.c: File not found或类似的东西)。

1 个答案:

答案 0 :(得分:2)

在@Michael的帮助下,我能够找到问题所在。

使用%ebp作为argv正如@Michael建议的那样(尽管他使用了%eax)。另一个问题是我需要将(%ebp)的值与0(空终止符)进行比较,然后在该点结束程序。

代码:

    movl 8(%esp), %ebp  /* Get argv.  */

pr_arg:
    cmpl $0, (%ebp)
    je endit

    pushl %ecx
    pushl (%ebp)
    pushl $output2
    call printf
    addl $8, %esp       /* remove output2 and current argument.  */
    addl $4, %ebp

    popl %ecx
    loop pr_arg

    ret