x86 lea指令
我试图很好地掌握 x86 中的 LEA 指令：
leal (%edx, %edx, 4), %eax
leal (%edx, %edx, 2), %eax
鉴于这两行，我知道：
eax = edx + edx*4
然后
eax = edx + edx*2 两个问题
。首先，如果这些指令像本例中那样按顺序出现，那么一旦第二行执行，eax寄存器就会被覆盖吗？到底什么会被加载到寄存器中？另一个地址？或者这是对这些寄存器指向的值进行算术运算？
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?
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是（两条指令后都会被覆盖）
EDX 描述的内存地址 + EDX 中存储为值的偏移量乘以 2
Yes (it is overwritten after both instructions)
The memory address described by EDX + the offset stored as value in EDX multiplied by 2