x86 64 assembly

Question

I am writing a self modifying code.

movq      $TARGET_CIA, 0x550(%rax)

This symbol TARGET_CIA is undefined initially and at run time I try to copy a 64 bit immediate value to this location. But at compile time this instruction takes the value of this undefined immediate value as 32 bit and when i try to copy the 64 bits, I see the signed extended 32 bits at its place. Is there a way to get this undefined symbol treated as 64 bit value?

Possible same: http://stackoverflow.com/questions/19415184/load-from-a-64-bit-address-into-other-register-than-rax — Ciro Santilli Путлер Капут 六四事, May 24 '15 at 21:05
Possible duplicate of [Difference between movq and movabsq in x86-64](https://stackoverflow.com/questions/40315803/difference-between-movq-and-movabsq-in-x86-64) — phuclv, Jul 04 '18 at 04:24

score 12 · Answer 1 · answered Apr 24 '11 at 19:49

12

You need

movabs $0x1234567890abcdef, 0x550(%rax)

The movabs instruction is required for 64 bit immediates.

answered Apr 24 '11 at 19:49

Gunther Piez

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Lol, just noticed this was an necro answer ;-) Anyway, maybe its of use for someone in some time ;-) – Gunther Piez Apr 24 '11 at 19:56

score 0 · Answer 2 · answered Nov 16 '10 at 04:38

0

I'd simply run another instruction to grab the second 32 bits. May not be the best way as I haven't done any ASM for a while, however it WILL work. :)

Good Luck Luke Peterson

answered Nov 16 '10 at 04:38

Luke Peterson

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x86 64 assembly

2 Answers2