“mov (%ebx,%eax,4),%eax"如何处理?工作? [英] How does "mov (%ebx,%eax,4),%eax" work?

问题描述

一直从事装配作业，并且在大多数情况下，我对装配非常了解.或者至少对于这项任务来说足够好.但是这个 mov 声明让我很不舒服.如果有人能解释一下这个 mov 语句是如何操作寄存器值的，我将不胜感激.

Been working on an assembly assignment, and for the most part I understand assembly pretty well. Or well at least well enough for this assignment. But this mov statement is tripping me up. I would really appreciate if someone could just explain how this mov statement is manipulating the register values.

mov (%ebx,%eax,4),%eax

mov (%ebx,%eax,4),%eax

附言我无法通过基本搜索找到这种特定类型的 mov 语句，因此如果我错过了它并重新提问，我深表歉意.

P.S. I wasnt able to find this specific type of mov statement by basic searches, so I appologize if I just missed it and am re asking questions.

推荐答案

AT&T程序集中完整的内存寻址模式格式为:

The complete memory addressing mode format in AT&T assembly is:

offset(base, index, width)

所以对于你的情况:

offset = 0
base = ebx
index = eax
width = 4

意思是指令是这样的:

eax = *(uint32_t *)((uint8_t *)ebx + eax * 4 + 0)

在类似 C 的伪代码中.

In a C-like pseudocode.

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