问题
Been working on an assembly assignment, and for the most part I understand assembly pretty well. Or well at least well enough for this assignment. But this mov statement is tripping me up. I would really appreciate if someone could just explain how this mov statement is manipulating the register values.
mov (%ebx,%eax,4),%eax
P.S. I wasnt able to find this specific type of mov statement by basic searches, so I appologize if I just missed it and am re asking questions.
回答1:
The complete memory addressing mode format in AT&T assembly is:
offset(base, index, width)
So for your case:
offset = 0
base = ebx
index = eax
width = 4
Meaning that the instruction is something like:
eax = *(uint32_t *)((uint8_t *)ebx + eax * 4 + 0)
In a C-like pseudocode.
来源：https://stackoverflow.com/questions/14900343/how-does-mov-ebx-eax-4-eax-work