在将字节存储到寄存器x86-64程序集中时出现问题 [英] Issue storing a byte into a register x86-64 assembly

问题描述

我正在尝试编写一个函数，该函数确定作为第一个参数给出的字符串的长度，因此％rdi将包含char * ptr.当我调用movb (%rdi),%rcx将％rdi指向的字符移动到％rcx时，出现以下错误:
incorrect register '%rdx' used with 'b' suffix 据我了解，只有某些寄存器可以在x86-64中保存一个字节，那么我可以使用哪一个将字节移入?还是我用来提取字符串中每个字节字符的方法不正确?

I am trying to write a function that determines the length of a string given as the first argument, so %rdi will contain char *ptr. When I call movb (%rdi),%rcx to move the character pointed to by %rdi into %rcx, I get the following error:
incorrect register '%rdx' used with 'b' suffix As I understand it, only certain registers can hold a byte in x86-64, so which ones can I use to move the byte into? Or is the method I am using to extract the character at each byte in the string incorrect?

推荐答案

所有通用寄存器具有低8位，分别可寻址为al，bl，cl，dl，sil ，dil，bpl，spl，r8b到r15b(英特尔文档使用后缀l).另外，还有一些8..15位也可寻址，即ah，bh，ch和dh.

All the general-purpose registers have the low 8 bits separately addressable as al, bl, cl, dl, sil, dil, bpl, spl, r8b through r15b (intel documentation uses l suffix). In addition, a few have bits 8..15 also addressable, namely ah, bh, ch and dh.

因此，如果您只想加载一个字节，则可以使用上述方法之一.另外，您可以使用零扩展或符号扩展来加宽字节数据，例如在movzbl (%rdi), %ecx的情况下(读取:将零扩展字节移动到long ).请注意，对32位寄存器进行操作会将父级" 64位寄存器的高32位清零，但是对8位或16位子寄存器进行操作将使其余位保持不变.

So if you only want to load a byte, you can use one of the above. Alternatively, you can use zero- or sign-extension to widen the byte data, for example in your case movzbl (%rdi), %ecx (read: move zero extended byte to long). Note that operating on 32 bit registers zeroes the top 32 bits of the "parent" 64 bit register, but operating on 8 or 16 bit sub-registers leaves the rest of the bits unchanged.

我觉得您应该(重新)阅读intel手册的基本体系结构部分.

I feel you should probably (re-)read the basic architecture section of the intel manuals.

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