x86 heap method

Question

Im trying to understand what this x86 instruction is doing:

 movl %eax, heap(,%rdx,4)

This is what I think its doing:

move value in eax to area in memory starting at label heap and displaced bytes from the value in rdx.

Is this correct?

Thanks

Answer 1

The AT&T syntax for memory operands is:

%segreg:disp(base,index,scale).

So the memory address in the instruction you've shown is (better represented in Intel syntax):

heap[rdx*4] 

So in other words, store the value in eax at rdx*4 bytes from the beginning of heap. From this, we can infer that heap is probably an array of 4-byte integers, and rdx is a loop counter or index into that array.

In Intel syntax, this would be:

mov heap[rdx*4], eax

For a comparsion between the syntaxes, see this page.