c - 汇编 EAX 寄存器无故复位

Question

我有以下汇编代码：

; File: strrev.asm
; A subroutine called from C programs.
; Parameters: string A
; Result: String is reversed and returned.


    SECTION .text
    global strrev
_strrev: nop
strrev:
    push    ebp
    mov ebp, esp

    ; registers ebx,esi, and edi must be saved if used
    push ebx
    push edi

    xor esi, esi    
    xor eax, eax
    mov ecx, [ebp+8]    ; load the start of the array into ecx
    jecxz   end     ; jump if [ecx] is zero
    mov edi, ecx

reverseLoop:
    cmp byte[edi], 0
    je  reverseLoop_1
    inc     edi 
    inc eax
    jmp reverseLoop



reverseLoop_1:

    mov esi, edi    ;move end of array into esi
    mov edi, ecx    ;reset start of array to edi

reverseLoop_2:
    mov al, [esi]
    mov bl, [edi]
    mov     [esi], bl
    mov [edi], al
    inc edi
    dec esi
    dec eax
    jnz reverseLoop_2

end:
    pop edi     ; restore registers
    pop ebx
    mov esp, ebp    ; take down stack frame
    pop ebp
    ret

在您开始循环通过 reverseLoop_2 之前，它工作正常。使用 gdb，eax 被列为 11，它应该是（这是我通过单独的 c 程序传入的字符串的长度）。这在调试器中显示为：

Breakpoint 2, reverseLoop_2 () at strrev.asm:40
40      mov al, [esi]
(gdb) display $eax
1: $eax = 11

但是，如果我将程序单步执行到下一行，它会重置为 0。

(gdb) next
41      mov bl, [edi]
1: $eax = 0

我需要保留 eax，因为它可以跟踪 reverseLoop_2 需要循环多少次。为什么调用mov后它会重置为0？

Answer 1

如果您eax用作循环计数器，则不应在循环内写入它：

reverseLoop_2:
  mov al, [esi]

请记住，这al是 的最低有效字节eax：

Answer 2

我认为这应该有效。

    mov eax, address of your string

    push esi
    push edi

    mov edi, eax
    mov esi, eax

    ; find end of string
    sub ecx, ecx
    not ecx
    sub al, al
    cld
    repne   scasb

    ; points to the byte after '0x00'
    dec edi
    dec edi

            ; main loop will swap the first with the last byte
            ; and increase/decrease the pointer until the cross each other

_loop:
    cmp esi, edi   ; if both pointers meet, we are done
    jg _done

    mov al, [edi]
    mov bl, [esi]
    mov [esi], al
    mov [edi], bl

    inc esi
    dec edi

    jmp _loop

_done:
    pop edi
    pop esi