在MASM程序集中逐字节循环和处理字符串 [英] Looping and processing string byte-by-byte in MASM assembly

问题描述

我正在使用MASM汇编，并且正在尝试编写一个逐字节处理字符串str1的循环，使用位操作将每个小写字母更改为相应的大写字母.如果这封信已经是大写字母，那就别说了.当我执行代码时，字符串str1似乎什么都没发生，并且我很难弄清为什么，也许我不应该这样处理我的数组，但是尽管如此，这还是代码:

I am using MASM assembly and I am trying to write a loop that processes the string str1 byte-by-byte, changing each lowercase letter into the corresponding capital letter using bit operations. If the letter is already capital, leave it alone. Nothing seems to happen to my string str1 when I execute my code and I'm having difficulty figuring out why, maybe I shouldn't be processing my array as such, but nonetheless, here's the code:

        .386
        .MODEL FLAT

        str1   dword  "aBcD", cr, Lf, 0
        ....

        .code
_start:
        output str1
        **sub esi, esi                 ; sum = 0
        lea ebx, str1
top:    mov al, [ebx + esi]            ; attempting to move each character value from              
                                       str1 into the register al for comparison and 
                                       possible conversion to uppercase
        add esi, 5
        cmp al, 0
        je zero
        sub al, 20h**                  ; convert lowercase to corresponding uppercase
        loop top
zero:   output zeromsg                 ; for TESTING of al purposes only
done:   output str1value
        output str1

没有任何变化，并且在转换未进行的最上面，它以相反的顺序打印字符串.为什么?打印为:"DcBa".任何查询将不胜感激！预先感谢.

Nothing changes , and on top of the conversion not taking place, the string it printing in reverse order. why? prints out as: "DcBa". Any inquiry would be appreciated! Thanks in advance.

推荐答案

您必须加载，处理字符并将其存储回去.您不存储它.

You must load the character, process it, and store it back. You don't store it.

类似的东西:

mov [esi+ebx], al

丢失.

您为什么从char中减去0x20?为什么将5加到esi?

Why do you sub 0x20 from the char? And why do you add 5 to esi?

更新

在开始编码之前，您应该考虑所需的步骤.

Before you start coding, you should think about what the required steps are.

1. 加载角色.
2. 如果字符为0，则表示字符串已完成.
3. 如果字符是大写字母，请将其转换
4. 存储字符
5. 前进到下一个字符并返回1

就是这样.现在，当您查看代码示例时，可以轻松地看到缺少的内容和出了问题的地方.

That's it. Now when you look at your code example, you can easily see what is missing and where you go wrong.

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