如何＆QUOT; MOV（％EBX，EAX％，4），％eax中＆QUOT;工作？ [英] How does "mov (%ebx,%eax,4),%eax" work?

问题描述

一直致力于在组装任务，并在大多数情况下我明白了装配pretty好。或者，嗯，至少不够好这项任务。但这个MOV语句被绊倒了我。我真的AP preciate如果有人可以只解释这MOV语句是如何操作的寄存器值。

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中

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

P.S。我不是能够找到这种特定类型由基本搜索MOV说法，所以我appologize如果我只是错过了它，我再次问问题。

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

完整的内存解决AT＆AMP模式格式;变速箱总成是：

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

offset(base, index, width)

因此​​，对于您的情况：

So for your case:

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

含义该指令是这样的：

Meaning that the instruction is something like:

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

在一个类似C伪code。

In a C-like pseudocode.

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