x86 cmpl and jne

Question

I'm tracing some x86 code for an assignment, and I was wondering what exactly "cmpl" does and how to predict whether or not the "jne" will be met.

80484bf:    83 7d f0 07             cmpl   $0x7,-0x10(%ebp)
80484c3:    75 16                   jne    80484db

BTW:This is AT&T Syntax. – Mackie Messer Mar 08 '13 at 08:09 — Mackie Messer, Mar 08 '13 at 08:09

score 16 · Answer 1 · edited Sep 11 '20 at 17:58

16

cmpl subtracts -0x10(%ebp) from $0x7 and modifies flags: AF CF OF PF SF ZF.

If memory at -0x10(%ebp) equals immediate 0x7 then the flag ZF is set. This is below EBP so it's probably a local variable, if this is an un-optimized build using EBP as a frame pointer.
jne 80484db means that if the two compared numbers are different (ZF=0), jump to 80484db

To summarize, your code is equivalent to :

compare A to 7
jump to 0x80484db if they are different.

edited Sep 11 '20 at 17:58

Peter Cordes

answered Mar 08 '13 at 08:04

Omar MEBARKI

Thank you very much. So ZF is set to 0 if -0x10(%ebp) minus $0x7 does not equal zero? Or is it the other way around? – Richarizard Mar 08 '13 at 08:42
2

You are welcome. Yes, ZF is set to 0 if -0x10(%ebp) minus $0x7 does not equal – Omar MEBARKI Mar 08 '13 at 08:54

1 Answers1