NASM x86_64 assembly in 32-bit mode: Why does this instruction produce RIP-Relative Addressing code?

Question

[bits 32]
    global _start

    section .data
    str_hello       db  \"HelloWorld\", 0xa
    str_hello_length    db      $-str_hello

    section .text

    _start         


        

Answer 1

The issue is probably that the offset of str_hello_length is greater than 32 bits. IA-32 doesn't support displacements of greater than 32 bits. The way around that is to use RIP-relative addressing, under the (often correct) assumption that the distance between the RIP and the address you're trying to reach fits in 32 bits. In this case, the base is RIP and the index is the instruction length, so if the instruction already has a base or an index, RIP-Relative can't be used.

Let's examine your various attempts:

str_hello_length equ $-str_hello
...
...
mov edx, str_hello_length

There's no memory access here, only a simply move with an immediate, so there's no addressing at all.

Next:

mov eax, str_hello_length       
mov edx, [eax]  ; count

Now the first instruction is a move with an immediate, which is still not a memory access. The second instruction has a memory access, but it uses eax as a base, and there's no displacement. RIP-relative is only relevant when there's a displacement, so there's no RIP-relative here.

Finally:

str_hello_length db $-str_hello
...
...
mov edx, [str_hello_length]     ; of course, without the brackets it'll load the address, which I don't want. I want the value stored at that address

Here you're using str_hello_length as your displacement. As I explained above, this will result in RIP-Relative addressing.