Linux 在装配中挣扎
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strlen in assembly
提问by Michel
I made my own implementation of strlen in assembly, but it doesn't return the correct value. It returns the string length + 4. Consequently. I don't see why.. and I hope any of you do...
Assembly source:
我在汇编中实现了自己的 strlen，但它没有返回正确的值。它返回字符串长度 + 4。因此。我不明白为什么......我希望你们中的任何人都这样做......
大会来源：
section .text
[GLOBAL stringlen:] ; C function
stringlen:
push ebp
mov ebp, esp ; setup the stack frame
mov ecx, [ebp+8]
xor eax, eax ; loop counter
startLoop:
xor edx, edx
mov edx, [ecx+eax]
inc eax
cmp edx, 0x0 ; null byte
jne startLoop
end:
pop ebp
ret
And the main routine:
和主要例程：
#include <stdio.h>
extern int stringlen(char *);
int main(void)
{
printf("%d", stringlen("h"));
return 0;
}
Thanks
谢谢
采纳答案by Jester
You are not accessing bytes (characters), but doublewords. So your code is not looking for a single terminating zero, it is looking for 4 consecutive zeroes. Note that won't always return correct value +4, it depends on what the memory after your string contains.
您访问的不是字节（字符），而是双字。因此，您的代码不是在寻找单个终止零，而是在寻找 4 个连续的零。请注意，不会总是返回正确的值 +4，这取决于字符串包含的内存。
To fix, you should use byte accesses, for example by changing edxto dl.
要修复，您应该使用字节访问，例如通过更改edx为dl.
回答by unwind
Not sure about the four, but it seems obvious it will always return the proper length + 1, since eaxis alwaysincreased, even if the first byte read from the string is zero.
不确定这四个，但似乎很明显它总是会返回正确的长度 + 1，因为eax它总是增加，即使从字符串中读取的第一个字节为零。
回答by Satya
I think your inc should be after the jne. I'm not familiar with this assembly, so I don't really know.
我认为你的公司应该在 jne 之后。我不熟悉这个程序集，所以我真的不知道。
回答by Zimbabao
Change the line
换线
mov edx, [ecx+eax]
to
到
mov dl, byte [ecx+eax]
and
和
cmp edx, 0x0 ; null byte
to
到
cmp dl, 0x0 ; null byte
Because you have to compare only byte at a time. Following is the code. Your original code got off-by-one error. For "h" it will return two h + null character.
因为您一次只能比较字节。以下是代码。您的原始代码出现了一个错误。对于“h”，它将返回两个 h + 空字符。
section .text
[GLOBAL stringlen:] ; C function
stringlen:
push ebp
mov ebp, esp ; setup the stack frame
mov ecx, [ebp+8]
xor eax, eax ; loop counter
startLoop:
xor dx, dx
mov dl, byte [ecx+eax]
inc eax
cmp dl, 0x0 ; null byte
jne startLoop
end:
pop ebp
ret
回答by Michel
Thanks for your answers. Under here working code for anyone who has the same problem as me.
感谢您的回答。在此处为与我有相同问题的任何人提供工作代码。
section .text
[GLOBAL stringlen:]
stringlen:
push ebp
mov ebp, esp
mov edx, [ebp+8] ; the string
xor eax, eax ; loop counter
jmp if
then:
inc eax
if:
mov cl, [edx+eax]
cmp cl, 0x0
jne then
end:
pop ebp
ret
回答by sharow
More easy way here(ASCII zero terminated string only):
这里更简单的方法（仅限 ASCII 零终止字符串）：
REPE SCAS m8
RPE SCAS m8
