IA32汇编代码到Y86汇编代码:Leal指令

问题描述

我正在研究如何将IA32汇编代码转换为Y86汇编代码，并且受制于以下IA32代码中的指令:

I am studying how to convert IA32 assembly code to Y86 assembly code, and I am stuck in the following instruction which is in IA32 code:

 leal(%edx, %eax), %eax

我找不到Y86代码的等效指令.我虽然有以下两个版本，但是我不确定哪个是正确的:

I cannot find the equivalent instructions for the Y86 code. I have though of two version as the following ones, but I am not sure which is right:

版本1:

 mrmovl (%edx), %ebx
 mrmovl (%eax), %esi
 addl %ebx, %esi
 rrmovl %esi, 5eax

版本2:

 addl %edx, %eax

有人有更好的主意吗?

推荐答案

LEA不访问内存，仅访问(地址)算术.因此，您的#2版本是正确的.

LEA doesn't access memory, it only does (address) arithmetic. As such your version #2 is correct.

请注意，在x86上，LEA不会影响标志，而ADD会影响标志. LEA还支持更复杂的有效地址语法，不过，将其转换为y86相当简单.例如，

Note that on x86 LEA doesn't affect flags, while ADD does. LEA also supports more complex effective address syntax, which is nevertheless quite straight-forward to transcribe to y86. For example,

leal offset(%eax, %ebx, 4), %edx

成为:

rrmovl %ebx, %edx
addl %edx, %edx
addl %edx, %edx
addl %eax, %edx
pushl %eax           # save eax which used as temporary for adding the offset
irmovl $offset, %eax
addl %eax, %edx
popl %eax            # restore eax

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