Addl instruction, x86

Question

You'll have to excuse me, I'm brand new to x86 assembly, and assembly in general.

So my question is, I have something like:

addl %edx,(%eax)

%eax is a register which holds a pointer to some integer. Let's call it xp

Does this mean that it's saying: *xp = *xp + %edx? (%edx is an integer)

I'm just confused where addl will store the result. If %eax is a pointer to an int, then (%eax) should be the actual value of that int. So would addl store the result of %edx+(%eax) in *xp? I would really love for someone to explain this to me!

I really appreciate any help!

Answer 1

Yes, this instruction is doing exactly what you think it's doing.

Most x86 arithmetic instructions take two operands: a source and a destination. In AT&T syntax (used here), the destination is always the right operand. So with an instruction like:

addl %edx, %eax

the values in edx and eax are added together and the result is stored in eax. However, in your example, memory indirect addressing is used (this is denoted by parentheses). This means that eax is treated as a pointer, so the right operand is taken from the address pointed to by eax, and the result is stored to the same address.