组件EAX寄存器无故重置
组件EAX寄存器无故重置
提问于 2013-05-06 20:48:25
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使用gdb，eax被列为11，这是应该的(这是我通过一个单独的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，如果我单步执行程序到下一行，它将重置为0。
(gdb) next
41 mov bl, [edi]
1: $eax = 0我需要保留eax，因为它记录了reverseLoop_2需要循环的次数。为什么在调用mov之后重置为0？
回答 2
Stack Overflow用户
发布于 2013-05-06 22:12:21
如果您使用eax作为循环计数器，则不应在循环内对其进行写入：
reverseLoop_2:
mov al, [esi]请记住，al是eax的最低有效字节：
Stack Overflow用户
发布于 2013-05-06 22:01:59
我想这应该行得通。
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/16399100
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