How to interpret this address -0x80(%rbp,%rax,4)

assembly x86-64 att
Theodros Zelleke picture Theodros Zelleke · Nov 13, 2012 · Viewed 9.1k times · Source

I'm currently trying to learn assembly language (and the effects of different compiler options) by analyzing simple C code snippets. Now I stumpled across the following instruction:

mov %edx,-0x80(%rbp,%rax,4)

What I do not understand is the expression for the target address -0x80(%rbp,%rax,4). The instruction assigns a value to a local array in a loop.

Answer

Jin Chen picture Jin Chen · Nov 13, 2012 
-0x80(%rbp,%rax,4) = *(%rbp + %rax * 4 + (-0x80))

So the following insruction:

mov %edx,-0x80(%rbp,%rax,4)

means let CPU move the value of register %edx to memory at address (%rbp + %rax * 4 + (-0x80)), this is AT&T-style assembly.

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