Memory indirect addressing movl - assembly

Question

I'm trying to understand how exactly memory indirect addressing works in assembly language with AT&T syntax.

movl (%eax), %ebx

movl %eax, (%ebx)

Here is a similar question that explains about memory indirect addressing

This is what I've understood:

In the first case, you load the data pointed to by the register %eax and store it in %ebx.

In the second case, you store the data in the register %eax to the address space pointed to by the register %ebx. Am I correct?

you are correct. In C it would be `ebx = *eax` and `*ebx = eax`. — ShinTakezou, Jun 10 '13 at 09:37

score 4 · Accepted Answer · answered Jun 10 '13 at 09:38

4

Basically the syntax is

movl source, destination

So movl (%eax), %ebx is indeed copy the value at address pointed to by %eax into %ebx. And movl %eax, (%ebx) is copy the value inside register %eax into the address pointed to by %ebx.

So indeed your understanding is correct.

answered Jun 10 '13 at 09:38

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Thanks!. Can look at the next part of the code with a better frame of mind now! – pistal Jun 10 '13 at 09:40

Memory indirect addressing movl - assembly

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