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From: Adam Jorgensen <adam.jorgensen.za@gmail.com>
Date: Wed, 24 Aug 2011 16:30:13 +0200
Subject: Re: Weird interaction with nested functions inside a decorator-producing function and closuring of outer data...
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Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:12144

Thanks :-) Sorry about the size, I wasn't sure what was relevant...

On 24 August 2011 15:29, Peter Otten <__peter__@web.de> wrote:
> Adam Jorgensen wrote:
>
>> Hi all, I'm experiencing a weird issue with closuring of parameters
>> and some nested functions I have inside two functions that
>> return decorators. I think it's best illustrated with the actual code:
>
> You should have made an effort to reduce its size
>> # This decorator doesn't work. For some reason python refuses to
>> closure the *decode_args parameter into the scope of the nested
>> decorate and decorate_with_rest_wrapper functions
>> # Renaming *decode_args has no effect
>> def rest_wrapper(*decode_args, **deco_kwargs):
>> =A0 =A0 def decorate(func):
>> =A0 =A0 =A0 =A0 argspec =3D getfullargspec(func)
>> =A0 =A0 =A0 =A0 decode_args =3D [argspec.args.index(decode_arg) for deco=
de_arg
>> in decode_args]
>
> I didn't read the whole thing, but:
>
>>>> def f(a):
> ... =A0 =A0 def g():
> ... =A0 =A0 =A0 =A0 =A0 =A0 a =3D a + 42
> ... =A0 =A0 =A0 =A0 =A0 =A0 return a
> ... =A0 =A0 return g
> ...
>>>> f(1)()
> Traceback (most recent call last):
> =A0File "<stdin>", line 1, in <module>
> =A0File "<stdin>", line 3, in g
> UnboundLocalError: local variable 'a' referenced before assignment
>
> Python treats variables as local if you assign a value to them anywhere i=
n
> the function. The fix is easy, just use another name:
>
>>>> def f(a):
> ... =A0 =A0 def g():
> ... =A0 =A0 =A0 =A0 =A0 =A0 b =3D a + 42
> ... =A0 =A0 =A0 =A0 =A0 =A0 return b
> ... =A0 =A0 return g
> ...
>>>> f(1)()
> 43
>
> In Python 3 you can also use the nonlocal statement.
>
> --
> http://mail.python.org/mailman/listinfo/python-list
>