Question

对于我的项目工作，我测试了一段代码片段。我发现由于堆栈的使用，此代码段出现了分段错误。但是，我必须使用堆栈解决它。任何人都可以帮我找出问题？我的程序将显示字符串变量的第二个字符。我的代码片段如下：

section.data
  string db "test",10,0
  msg2   db "%c",10,0
main:

xor eax,eax
mov eax, string
mov ebx,1    ; suppose I want to get the second character of string. SO I store the index of that character in ebx

pusha
push eax
push ebx
add esp,8
popa

pop dword[ebx]   ;  **I assume that this following 3 lines arise segmentation fault**   
pop eax          ;
mov bl,[eax+ebx] ; I want to move the fisrt character to BL using the index value stored in ebx which i popped just.

pusha
push ebx
call putchar
add esp,4
popa

pusha
push msg2
call printf
add esp,4
popa

在此，为了您的善意考虑，我想明确指出，此代码段的目的是了解如何操作堆栈。

在这里，@ nrz最近给了我一个关于以下代码的想法，我在这里编辑了上面的代码：

  section.data
    string db "test",10,0
    msg2   db "%c",10,0
  main:
    xor eax,eax
    mov eax, string
    mov ebx,1  ; suppose I want to get the second character of string. SO I store the index of that character in ebx
    mov   eax,string
    movzx eax,byte [eax]
    push  eax        ; these push and pop are for solving it using the stack,
    pop   ebx
    pusha
    push ebx
    call putchar
    add esp,4
    popa
    pusha
    push msg2
    call printf
    add esp,4
    popa

我的查询是特定的：

我会给出索引值。它应该在ebx寄存器中吗？
最重要的是，我使用堆栈的主要思想是使用我之前在string中推送的索引值来处理ebx变量的每个字符。 [这是强制性的。有可能吗？]
我想将输出存储在一个8位寄存器中。

所以我的所有想法都是这样的：
```
mov al, [string+ebx]   ;is it possible?
```
我必须从堆栈中获取ebx的值。我将在ebx中设置一个值，然后在push ebx中设置一个值，在mov al,[string+ebx]时，我将pop ebx获取mov指令的值。更有可能的说明如下：
```
 pop ebx  
 mov al,[string+dword[ebx]]  ;which is a wrong statement shown by NASM
```

我急切地等待你的回应。

感谢你，

Answer 1

代码中存在一些错误和不必要的指示：

mov eax, string  ; this line makes eax point to string, but...
xor eax,eax      ; this line zeroes eax immediately. delete this line.
mov ebx,1        ; this line is unnecessary, but does not cause problems.

pusha            ; unnecessary.
push eax         ; unnecessary.
push ebx         ; unnecessary.
add esp,8        ; unnecessary.
popa             ; unnecessary.

pop dword[ebx]   ; [1] = [esp], esp = esp+4, probably causes segmentation fault.
pop eax          ; eax = [esp], esp = esp+4
mov bl,[eax+ebx]

编辑：打印所有字符，只需遍历字符串：

section.data
  string db "test",10
  string_length equ $-string ; number of characters in the string.
                             ; no zero needed in the end of the string.
  msg2   db "%c",10,0
main:
    xor eax,eax      ; eax is the index to the string.

char_loop:
    push dword [eax+string]      ; push 4 first characters into stack.
    and  dword [esp], 0x000000ff ; x86 is little-endian, so 1st character
                                 ; is the lowest byte.
    pop  ebx                     ; pop the ASCII code of the character into ebx.

    pushad
    push ebx
    call putchar
    add esp,4
    popad

    inc eax
    cmp eax,string_length
    jb  char_loop

修改：使用ebx作为索引来匹配问题中的更改的新代码。

section.data
  string db "test",10
  string_length equ $-string ; number of characters in the string.
                             ; no zero needed in the end of the string.
  msg2   db "%c",10,0
main:
    xor  eax,eax             ; zero eax, for putchar.
    xor  ebx,ebx             ; ebx is the index to the string.

char_loop:
    ;push dword [ebx+string]      ; push 4 first characters into stack.
    ;and  dword [esp], 0x000000ff ; x86 is little-endian, so 1st character
    ;                             ; is the lowest byte.
    ;pop  eax                     ; pop the ASCII code of the character into ebx.

    mov   al,[ebx+string]         ; this is the normal way to do it.
                                  ; the above 3 instructions do it "using stack",
                                  ; as requested in the question.
    pushad
    push eax
    call putchar
    add esp,4
    popad

    inc ebx
    cmp ebx,string_length
    jb  char_loop

如何在Ubuntu 11.04中使用NASM维护堆栈

1 个答案: