scope of function parameters

[Steven D'Aprano]

On Mon, 30 May 2011 04:38:26 +1000, Chris Angelico wrote:

> On Mon, May 30, 2011 at 4:01 AM, Chris Rebert <clp2 at rebertia.com> wrote:
>> def foo():
>>    print bar
>>    bar = 42
>>
>> foo()
>>
>> ===>
>> Traceback (most recent call last):
>>  File "<stdin>", line 1, in <module>
>>  File "<stdin>", line 2, in foo
>> UnboundLocalError: local variable 'bar' referenced before assignment
> 
> Wow
> 
> I thought it basically functioned top-down. You get a different error on
> the print line if there's a "bar = 42" *after* it. This could make
> debugging quite confusing.

UnboundLocalError is a subclass of NameError, so it will still be caught 
by try...except NameError.

If you're crazy enough to be catching NameError :)

Go back to Python1.5, and there was no UnboundLocalError. It was 
introduced because people were confused when they got a NameError after 
forgetting to declare something global:


>>> def f():
...     print a
...     a = a + 1
...
>>> a = 42
>>> f()
Traceback (innermost last):
  File "<stdin>", line 1, in ?
  File "<stdin>", line 2, in f
NameError: a


While UnboundLocalError is jargon, and not the easiest error message to 
comprehend, at least it confuses in a different way :)


-- 
Steven