x86简单mov指令 [英] x86 simple mov instruction

问题描述

这是一个简单的问题，但我在Google上找不到可靠的答案.

This is a simple question but I can't find reliable answers on google.

此说明是什么意思:

movl %eax, (%esi, %ecx, 4)

是否会将寄存器eax中的值移动到(%esi, %ecx, 4)指向的内存中的值?

Is it move the value at register eax to the value in memory that (%esi, %ecx, 4) is pointing too?

(%esi, %ecx, 4)用于数组.因此，这意味着Array [Xs + 4i]，其中Xs是Array内存中的起点，而i只是整数数组中的偏移量.

(%esi, %ecx, 4) is for an array. So it means Array[Xs + 4i] where Xs is the starting point in memory for the Array and i is just an offset in the integer array.

推荐答案

完全正确.这是AT& T语法，因此源首先出现，然后是目的地.因此，它将eax寄存器的内容存储到存储位置esi + 4*ecx.

Exactly correct. This is AT&T syntax, so the source comes first, then the destination. Thus, it stores the contents of the eax register to the memory location esi + 4*ecx.

如果您想将其视为一个数组，它会将eax存储到基于esi的4字节对象数组的第ecx个条目.

If you like to think of this as an array, it stores eax to the ecxth entry of an array of 4-byte objects based at esi.

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