我正在尝试编写一个汇编程序，它调用c中的函数，该函数将使用预定义字符替换字符串中的某些字符，前提是char数组中的当前字符符合某些条件。
我的c档案：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
//display *((char *) $edi)
// These functions will be implemented in assembly:
//
int strrepl(char *str, int c, int (* isinsubset) (int c) ) ;
int isvowel (int c) {
if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u')
return 1 ;
if (c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U')
return 1 ;
return 0 ;
}
int main(){
char *str1;
int r;
// I ran my code through a debugger again, and it seems that when displaying
// the character stored in ecx is listed as "A" (correct) right before the call
// to "add ecx, 1" at which point ecx somehow resets to 0 when it should be "B"
str1 = strdup("ABC 123 779 Hello World") ;
r = strrepl(str1, '#', &isdigit) ;
printf("str1 = \"%s\"\n", str1) ;
printf("%d chararcters were replaced\n", r) ;
free(str1) ;
return 0;
}
我的.asm文件：
; File: strrepl.asm
; Implements a C function with the prototype:
;
; int strrepl(char *str, int c, int (* isinsubset) (int c) ) ;
;
;
; Result: chars in string are replaced with the replacement character and string is returned.
SECTION .text
global strrepl
_strrepl: nop
strrepl:
push ebp ; set up stack frame
mov ebp, esp
push esi ; save registers
push ebx
xor eax, eax
mov ecx, [ebp + 8] ;load string (char array) into ecx
jecxz end ;jump if [ecx] is zero
mov esi, [ebp + 12] ;move the replacement character into esi
mov edx, [ebp + 16] ;move function pointer into edx
xor bl, bl ;bl will be our counter
firstLoop:
add bl, 1 ;inc bl would work too
add ecx, 1
mov eax, [ecx]
cmp eax, 0
jz end
push eax ; parameter for (*isinsubset)
;BREAK
call edx ; execute (*isinsubset)
add esp, 4 ; "pop off" the parameter
mov ebx, eax ; store return value
end:
pop ebx ; restore registers
pop esi
mov esp, ebp ; take down stack frame
pop ebp
ret
当通过gdb运行并在断点处设置断点时; BREAK，在我采取步骤调试命令后出现以下错误：
Program received signal SIGSEGV, Segmentation fault.
0x0081320f in isdigit () from /lib/libc.so.6
isdigit是我在c文件中包含的标准c库的一部分，所以我不知道该怎么做。
编辑：我已经编辑了我的firstLoop并包含了一个secondLoop，它应该用“＃”替换任何数字，但它似乎取代了整个数组。
firstLoop:
xor eax, eax
mov edi, [ecx]
cmp edi, 0
jz end
mov edi, ecx ; save array
movzx eax, byte [ecx] ;load single byte into eax
mov ebp, edx ; save function pointer
push eax ; parameter for (*isinsubset)
call edx ; execute (*isinsubset)
;cmp eax, 0
;jne end
mov ecx, edi ; restore array
cmp eax, 0
jne secondLoop
mov edx, ebp ; restore function pointer
add esp, 4 ; "pop off" the parameter
mov ebx, eax ; store return value
add ecx, 1
jmp firstLoop
secondLoop:
mov [ecx], esi
mov edx, ebp
add esp, 4
mov ebx, eax
add ecx, 1
jmp firstLoop
使用gdb，当代码进入secondloop时，一切都是正确的。 ecx显示为“1”，这是从.c文件传入的字符串中的第一个数字。 Esi正在显示为“＃”。然而，在我做了[ecx]后，esi似乎崩溃了。 ecx在这一点上显示为“＃”，但是一旦我增加1来到达数组中的下一个字符，它就会被显示为“/ 000”。 1后面的每个字符都替换为“＃”，显示为“/ 000”。在我让secondLoop尝试用“＃”替换字符之前，我只是用自己的firstLoop循环来查看它是否可以通过整个数组而不会崩溃。确实如此，并且每次增量后ecx都显示为正确的字符。我不确定为什么做mov [ecx]，esi会将ecx的其余部分设置为null。
答案 0 :(得分：1)
在firstLoop:中，您使用以下字符加载字符串中的字符：
mov eax, [ecx]
在一个平局而不是一个字节上加载4个字节。因此，您传递给int的{{1}}可能远远超出其处理范围（它可能使用简单的表查找）。
您可以使用以下Intel asm语法加载单个字节：
isdigit()其他一些事情：
movzx eax, byte ptr [ecx]
答案 1 :(得分：0)
我在您的代码中添加了一些注释： -
; this is OK: setting up the stack frame and saving important register
; on Win32, the registers that need saving are: esi, edi and ebx
; the rest can be used without needing to preserve them
push ebp
mov ebp, esp
push esi
push ebx
xor eax, eax
mov ecx, [ebp + 8]
; you said that this checked [ecx] for zero, but I think you've just written
; that wrong, this checks the value of ecx for zero, the [reg] form usually indicates
; the value at the address defined by reg
; so this is effectively doing a null pointer check (which is good)
jecxz end
mov esi, [ebp + 12]
mov edx, [ebp + 16]
xor bl, bl
firstLoop:
add bl, 1
; you increment ecx before loading the first character, this means
; that the function ignores the first character of the string
; and will therefore produce an incorrect result if the string
; starts with a character that needs replacing
add ecx, 1
; characters are 8 bit, not 32 bit (mentioned in comments elsewhere)
mov eax, [ecx]
cmp eax, 0
jz end
push eax
; possibly segfaults due to character out of range
; also, as mentioned elsewhere, the function you call here must conform to the
; the standard calling convention of the system (e.g, preserve esi, edi and ebx for
; Win32 systems), so eax, ecx and edx can change, so next time you call
; [edx] it might be referencing random memory
; either save edx on the stack (push before pushing parameters, pop after add esp)
; or just load edx with [ebp+16] here instead of at the start
call edx
add esp, 4
mov ebx, eax
; more functionality required here!
end:
; restore important values, etc
pop ebx
pop esi
mov esp, ebp
pop ebp
; the result of the function should be in eax, but that's not set up properly yet
ret
关于内循环的评论： -
firstLoop:
xor eax, eax
; you're loading a 32 bit value and checking for zero,
; strings are terminated with a null character, an 8 bit value,
; not a 32 bit value, so you're reading past the end of the string
; so this is unlikely to correctly test the end of string
mov edi, [ecx]
cmp edi, 0
jz end
mov edi, ecx ; save array
movzx eax, byte [ecx] ;load single byte into eax
; you need to keep ebp! its value must be saved (at the end,
; you do a mov esp,ebp)
mov ebp, edx ; save function pointer
push eax ; parameter for (*isinsubset)
call edx ; execute (*isinsubset)
mov ecx, edi ; restore array
cmp eax, 0
jne secondLoop
mov edx, ebp ; restore function pointer
add esp, 4 ; "pop off" the parameter
mov ebx, eax ; store return value
add ecx, 1
jmp firstLoop
secondLoop:
; again, your accessing the string using a 32 bit value, not an 8 bit value
; so you're replacing the matched character and the three next characters
; with the new value
; the upper 24 bits are probably zero so the loop will terminate on the
; next character
; also, the function seems to be returning a count of characters replaced,
; but you're not recording the fact that characters have been replaced
mov [ecx], esi
mov edx, ebp
add esp, 4
mov ebx, eax
add ecx, 1
jmp firstLoop
您似乎遇到了内存工作方式的问题，您在8位和32位内存访问之间感到困惑。