[Python-ideas] Tweaking closures and lexical scoping to include the function being defined

[Ron Adam]

On Thu, 2011-09-29 at 13:01 -0700, Guido van Rossum wrote:
> On Thu, Sep 29, 2011 at 12:03 PM, Eric Snow <ericsnowcurrently at gmail.com> wrote:
> > The alternative is to leave nonlocal as just a simple statement, but
> > change its behavior when the name is not found inside a containing
> > function scope.  Currently that is a syntax error.
> 
> For a reason. It would be too easy for a typo to produce the wrong
> interpretation.

But this is exactly how it works now!  The behavior of nonlocal doesn't
need to be changed.  It already behaves that way with closures. :-)


Cheers,
   Ron

(PS... Pleas repost if this doesn't show up on the python-ideas.  My
emails aren't getting there.  My apologies for duplicates.)



Python 3.2rc2+ (py3k:88358, Feb  6 2011, 08:56:00) 
[GCC 4.4.5] on linux2
Type "help", "copyright", "credits" or "license" for more information.


Closure as defaults.

>>> def a(x):
...    def b(y):
...       return x+y
...    return b
... 
>>> c = a(5)
>>> c(4)
9
>>> c(5)
10


Closures with nonlocal.

>>> def a(x):
...     def b(y):
...         nonlocal x
...         x = x + y
...         return x
...     return b
... 
>>> c = a(3)
>>> c(1)
4
>>> c(2)
6
>>> c(3)
9


Closure won't allow changes to cells without nonlocal. 

>>> def a(x):
...     def b(y):
...         x = x + y
...         return x
...     return b
... 
>>> c = a(3)
>>> c(1)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in b
UnboundLocalError: local variable 'x' referenced before assignment
>>>