Georg Brandl wrote:
> Marcin 'Qrczak' Kowalczyk wrote:
>> The rule should be:
>>
>> The keyword 'nonlocal' causes the lookup to be performed as if there
>> were no assignments to that variable in the scope containing the
>> 'nonlocal' declaration.
>
> Plus, if there's no binding in an enclosing scope, an error is raised.
>
> (Which brings up the assymetry to today's global again, but is that really
> a problem?)
I'd narrow the requirement such that module globals don't count - if you
declare a variable nonlocal it *must* be initialised in an enclosing function
scope or you will get a syntax error similar to this one:
>>> def f():
... x = 1
... del x
... def g():
... return x
... return g
...
SyntaxError: can not delete variable 'x' referenced in nested scope
Here, the compiler knows that g() references back to 'x' and hence considers
the attempt to delete 'x' nonsensical. Attempting to reference 'nonlocal x'
when x isn't a closure variable is also nonsensical - if you want a local
variable then remove the declaration, if you want a global then declare x as a
global.
Cheers,
Nick.
--
Nick Coghlan | [EMAIL PROTECTED] | Brisbane, Australia
---------------------------------------------------------------
http://www.boredomandlaziness.org
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