How does "mov (%ebx,%eax,4),%eax" work?

assembly att addressing-mode
Scalahansolo picture Scalahansolo · Feb 15, 2013 · Viewed 7.3k times · Source

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.

Answer

Carl Norum picture Carl Norum · Feb 15, 2013

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.

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