How is the x86 JAE instruction related to the carry flag?

Question

I have some x86 code which looks like:

;  The carry flag is set to 1 here
jae    an_address  ; The jump instruction does not take place

Does this make sense?

I thought the jump should take place because 1 is greater than or equal to 0, the definition of JAE?

Solution 1

jae is the same as jnc, i.e. jump if CF == 0. Choice between all 3 mnemonics (including jnb) is up to programmer. CF isn't set here by mov but by a previous instruction. The mnemonics jae is recommended after compare instruction (cmp) which does subtraction. You can get more details in Intel or AMD software developer manuals.

Solution 2

jae means Jump if above or equal. It will jump if the carry flag is equal to 0.

You're looking for jnae or jb

Comments

[user997112]

I'm not looking for anything :) I am looking at code which contains jae, CF is zero and the jump is not being taken?

[Nowayz]

You must be looking at something wrong, becasue that's not how jae behaves

[user997112]

Im debugging in Visual Studio, I can see the value of the carry flag and when I step-in, the debugger doesn't jump to the address of the branch, just continues right over it.

[Nowayz]

in your question you say the carry flag is 1, is this not the case?

[user997112]

Sorry- I meant carry is initially zero, then it is set to 1, then the JAE occurs and there's no branch.

[Rudy Velthuis]

Because JAE jumps when the carry flag is 0, so if it is not 0, the jump should not take place, as you observe.