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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.
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汇编指令mov寄存器,[寄存器] [寄存器]
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