Quote from: lumen on January 26, 2013, 11:48:47 PM
Actually no, I made it especially to fit your EXACT rules.

Lumen, you haven't been able to understand almost anything I've posted.   Changing that now would only be refreshing.

Quote from: sarkeizen on January 26, 2013, 11:54:49 PM
Lumen, you haven't been able to understand almost anything I've posted.   Changing that now would only be refreshing.

Really?
Are you now wondering where you went wrong?
To fix the problem is very simple, but you don't yet get it?
Do you want clues or should I just come out and solve it for you.

Quote from: lumen on January 27, 2013, 12:06:39 AM
Really?
Are you now wondering where you went wrong?
To fix the problem is very simple, but you don't yet get it?
Do you want clues or should I just come out and solve it for you.

This is a good example about what I mean concerning your inability to understand what I say.

Quote from: sarkeizen on January 27, 2013, 12:11:02 AM
This is a good example about what I mean concerning your inability to understand what I say.

Wow, setting up some defense already.

Ok, here it is then.

Because the ONLY case that causes a problem is by running itself with a pointer back to itself. This would return an invalid false becaus it would see an endless loop.

Since there is only one case like this we can simply hardcode the result.
The very first instructions simply run a byte by byte comparison of the program it's going to run and itself.
If it finds it is indeed itself, it does not run it, it simply returns TRUE, because it already knows this program ends.
Now you can run this EXACTLY as you have shown to cause the return error, and now ALL cases work and return the correct result.
Ok , now you can claim you don't understand what I just did. 

Quote from: lumen on January 27, 2013, 12:26:04 AM
If it finds it is indeed itself, it does not run it, it simply returns TRUE, because it already knows this program ends.

What is "it" in this sentence.  You used it six times.

Please replace it with either WillThisEnd() or WillThisEndOnSelf() where applicable.