汇编指令mov寄存器，[寄存器] [寄存器]

本文介绍了汇编指令mov寄存器，[寄存器] [寄存器]的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在理论上在高中学习ASM 8086(MASM，x86).

I'm studying ASM 8086 theoretically on highschool (MASM, x86).

.data var dd 421,422, 443, 442, 444, 217, 432 .code ; some code mov esi, (OFFSET var)+4 mov ebx, 4 mov edx, [ebx][esi] ; that's the line I don't uderstand

我运行了该程序，之后EDX == 000001BBh == 443 该代码中最后一行的含义是什么?它是做什么的?

I ran that program and after that EDX == 000001BBh == 443 What's the meaning of last line in that code? What does it do?

推荐答案

esi指向var后4个字节，即422. ebx是4.

esi points 4 bytes after var, which is 422. ebx is 4.

[ebx][esi]表示[ebx+esi]，[]是指针运算符.

[ebx][esi] is something which denotes [ebx+esi] and the [] is a pointer operator.

所有这些都将使[ebx][esi]点比422还要远4个字节，并且显然可以在其中找到443.

All this together will make [ebx][esi] point yet 4 bytes farther than 422 and obviously 443 can be found there.