Tim Rau wrote:
> I'm sorry: I forgot to say what my problem was. Python seems to think
> that nextID is a local, and complains that it can't find it. THis is
> not the full text of the function, just the level that is causing
> errors. the lack of : on the if is a transcription error.
> Traceback (most recent call last):
> File "C:\Documents and Settings\Owner\My Documents\NIm's code\sandbox
> \sandbox.py", line 248, in <module>
> thing.step()
> File "C:\Documents and Settings\Owner\My Documents\NIm's code\sandbox
> \sandbox.py", line 112, in step
> allThings[len(allThings)-1].id = nextID
> UnboundLocalError: local variable 'nextID' referenced before
> assignment
But in the next line in the same function you say
nextID+= 1
Since there is an assignment to nextID in the same function without
a global statement, you tell Python, that nextID is a local variable.
Note: Assignment (anywhere in a function) is a local definition unless
there is a global statement in this function
> I want to know why it says 'local variable' it never had a problem
> when just allThings was in there.
>
> as for the objection to global variable: If I didn't make them global,
> I'd make them part of a singleton class called gameVars that would get
> passed everywhere. It's unlikely that they'll get mixed in with
> anyhting else, as they fulfill a unique function. Also, I think it's
> more convenient, and I am, after all, my own employer when it comes to
> programming.
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
--
http://mail.python.org/mailman/listinfo/python-list
