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

Question

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 1

The machine command will copy the content of %edx to the address given by %rbp + 4 * %rax - 0x80. It seems %rax is holding the index to that array and %rbp - 0x80 is the base address.

Take a look here to get a better understanding for the AT&T syntax.

Answer 2

-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.