从寄存器打印字符 [英] Printing character from register

问题描述

我使用的是NASM汇编器。

I'm using the NASM assembler.

返回eax寄存器的值应该是一个字符，当我试图打印整数表示时，它的a值看起来像一个内存地址。我期待的字母的十进制表示。例如，如果字符'a'被移动到eax，我应该看到97被打印（'a'的十进制表示）。但不是这样。

The value returned to the eax register is supposed to be a character, when I attempt to print the integer representation its a value that looks like a memory address. I was expecting the decimal representation of the letter. For example, if character 'a' was moved to eax I should see 97 being printed (the decimal representation of 'a'). But this is not the case.

section .data
             int_format db "%d", 0
;-----------------------------------------------------------

mov eax, dword[ebx + edx]
push eax
push dword int_format
call _printf   ;prints a strange number
add esp, 8

xor eax, eax
mov eax, dword[ebx + edx]
push eax
call _putchar ;prints the correct character!
add esp, 4

最终我想比较字符，所以重要的是eax获得正确的字符的十进制表示。

So what gives here? ultimately I want to compare the character so it is important that eax gets the correct decimal representation of the character.

推荐答案

mov eax, dword[ebx + edx]

32位）从指向ebx + edx的地址。如果你想要一个单一的字符，你需要加载一个字节。为此，你可以使用movzx指令：

You are loading a dword (32 bits) from the address pointed to ebx+edx. If you want a single character, you need to load a byte. For that, you can use movzx instruction:

movzx eax, byte[ebx + edx]

这将加载一个字节到 eax

另一种选择是在加载完成后屏蔽掉额外的字节，如下所示：c $ c> al dword，例如：

Another option would be to mask out the extra bytes after loading the dword, e.g.:

and eax, 0ffh

或

movxz eax, al

对于 putchar ，它工作，因为它将传递的值解释为char，寄存器中的高三个字节，只考虑低字节。

As for putchar, it works because it interprets the passed value as char, i.e. it ignores the high three bytes present in the register and considers only the low byte.

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