86 lea指令 [英] x86 lea instruction

问题描述

我想获得在x86的机关指令良好的抓地力：

 莱亚尔（％EDX，EDX％，4），EAX％
莱亚尔（％EDX，EDX％，2），％eax中
 

由于这两条线，我知道：

EAX = EDX + EDX * 4

然后

EAX = EDX + EDX * 2

两个问题。首先，如果这些指示出现在序列在本实施例中，EAX寄存器一旦第二线执行覆盖？而究竟会被加载到寄存器？另一个地址？或者是对这些寄存器指向值这个算术运算？

解决方案

  

如果这些指示出现在序列在本实施例中，eax中
  寄存器是重写一次，第二行执行？

是（这是两个指令后覆盖）

  

和究竟会被加载到寄存器？另一个地址？

通过EDX描述的内存地址+偏移量存储为乘以2 EDX值

I am trying to get a good grip on the LEA instruction in x86:

leal (%edx, %edx, 4), %eax
leal (%edx, %edx, 2), %eax

Given these two lines, i know that:

eax = edx + edx*4

and then

eax = edx + edx*2

Two questions. First, if these instructions appear in sequence as in this example, the eax register is overwritten once the second line executes? And what exactly would be loaded into the register? Another address? Or is this doing arithmetic on the values that these registers point to?

解决方案

if these instructions appear in sequence as in this example, the eax register is overwritten once the second line executes?

Yes (it is overwritten after both instructions)

And what exactly would be loaded into the register? Another address?

The memory address described by EDX + the offset stored as value in EDX multiplied by 2

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