How is $example different from example(%rip)?

Question

I've spent 2 hours googling, but to no avail --- there are not many beginner-level guides for assembly, and the course I am taking right now dos not do a very good job at explaining some stuff.

Anyway; I've been trying to work with SSE and tried comparing two doubles using comisd instruction. I've spent a lot of time to understand how to hard-code a non-integer constant (let's call it example, it is declared as

example:
    .long   3794832442
    .long   1044740494

); but after I've done that, I couldn't get it to work --- "operand type mismatch". I've turned to C-to-assembly translator and found out that instead of my comisd $example, %xmm0 it used comisd example(%rip), %xmm0, and it turned out to work. Now I don't understand, how does it work and how are these different?

Answer 1

$example is simply the absolute address of that variable - it doesn't mean a memory access.

You get the "operand type mismatch" because comisd does not support an immediate operand. comisd example, %xmm0 (without the $) would have worked, as that is a memory reference.

example(%rip) is using the PC-relative addressing mode introduced in x86-64, which makes the code position-independent because instead of using an absolute address, it uses an offset from the current instruction pointer.

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_{Based on an answer of Jonathon Reinhart.}