ARM 程序集 - 替换字符串中字符的代码 [英] ARM assembly - code to replace character on a string

问题描述

我有这个 C 驱动程序

I have this C driver program

#include <stdlib.h>
#include <stdio.h>

extern void subs( char *string, char this_c, char that_cr ) ;

int main(int argc, char *argv[] )
{
    char this_c= 'e' ;  
    char that_c = 'X' ;
    char orgstr[] = "sir sid easily teases sea sick seals" ;

    subs( orgstr, this_c, that_c ) ;
    printf( "Changed string: %s\n", orgstr ) ;

    exit( 0 ) ;
}

我必须制作一个将字符串上的e"更改为x"的 arm 程序，到目前为止，这就是我所拥有的 \

I have to make an arm program that changes the 'e' on the String to 'x', so far this is what I have \

.global subs

subs:   
    stmfd sp!, {v1-v6, lr} //string entry
    mov v1, #0 //set index to 0
    mov v2, #0 // set count to 0


loop :
    ldrb    v3, [a1,v1] // get the first char
    cmp     a2,v3    //compare char if its the same to the one that has to b chang
    mov v3, a3   // change the character for the new character
    addeq   v2, v2, #1 //increment count
    add     v1, v1, #1 // increment index
    cmp v3,#0   //end of string
    bne     loop
    mov     a1,v2 //return value

    ldmfd   sp!, {v1-v6, pc}
.end

这给了我一个无限循环，我被卡住了，谁能帮我找出问题所在??

This is giving me an infinite loop though and I'm stuck, can anyone help me figure out where the problem is??

这就是我目前所拥有的，

So this is what i have so far,

.global subs

subs:   
    stmfd sp!, {v1-v6, lr} //string entry
    mov v1, #0 //set index to 0
    mov v2, #0 // set count to 0


loop :
    ldrb    v3, [a1,v1] // get the first char
    cmp     a2,v3    //compare char if its the same to the one that has to b chang
    moveq   v3, a3   // change the character for the new character
    addeq   v2, v2, #1 //increment count
    add     v1, v1, #1 // increment index
    cmp v3,#0   //end of string
    bne     loop
    mov     a1,v2 //return value

    ldmfd   sp!, {v1-v6, pc}
.end

它运行但字符从未改变...，输出和结尾与输入相同，出于某种原因，a2注册表为空...

it runs but the character is never changed..., the output and the end is the same one as the input, for some reason, the a2 registry is null...

推荐答案

这行完全错误

mov v3, a3   // change the character for the new character

这应该是一个条件存储回到字符串中.

This should be a conditional store back into the string.

streqb a3, [a1, v1]

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