namespaces & indentation
sholden at holdenweb.com
Fri Dec 22 17:57:25 EST 2000
Chris Schaller <Chris.Schaller at web.de> wrote in message
news:920f54$m2f$1 at news.mch.sbs.de...
> Dear Pythoners...
> I always thought that I understood Python's scoping, but then I came
> that code:
> > def t(x):
> > if x == 1:
> > c = 'yeah'
> > return c
> If I execute t(1), I'll get 'yeah', but otherwise there'll be an error
> message. According to the indentation level c is only valid within the
> if-clause, but not on the outside. So I guess, the indentation level only
> defines the lines of code that belong to the if-clause, but has nothing to
> with scopes.
The indentation level has nothgin to do with scopes! The scope of c is the
function definition, because c is bound (assigned to) within the function
body. The interpreter therefore expects to find c in the local namespace.
So, when x == 1 you return the value 'yeah' because that's what you assigned
When x != 1 you get an error, because the interpreter (having told itself
during compilation that c is local to the function) can't find c in the
function's namespace, because you haven't bound a value to it!
> That's very confusing, I've already programmed a lot in Python, but only
> due to a program error (I forgot to initialize c at the beginning of the
> function - a common error) and an unsuccessful test this happened to a
> after months of error-free running.
> Any comments?
I think you understand the namespace rules quite well, you just hadn't
previously realized that names are created dynamically on assignment rather
than statically during compilation. If you haven't assigned a value to them
you get an UnboundLocalError exception (which describes the problem quite
succinctly, don't you think?).
Has this helped?
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