Creating a class within a function and access a function defined in the containing function’s scope [duplicate]

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That’s an artifact of Python’s name resolution rules: you only have access to the global and the local scopes, but not to the scopes in-between, e.g. not to your immediate outer scope.

EDIT: The above was poorly worded, you do have access to the variables defined in outer scopes, but by doing x = x or mymethod = mymethod from a non-global namespace, you’re actually masking the outer variable with the one you’re defining locally.

In example 2, your immediate outer scope is the global scope, so MyClass can see mymethod, but in example 4 your immediate outer scope is my_defining_func(), so it can’t, because the outer definition of mymethod is already masked by its local definition.

See PEP 3104 for more details about nonlocal name resolution.

Also note that, for the reasons explained above, I can’t get example 3 to run under either Python 2.6.5 or 3.1.2:

>>> def myfunc():
...     x = 3
...     class MyClass(object):
...         x = x
...     return MyClass
... 
>>> myfunc().x
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in myfunc
  File "<stdin>", line 4, in MyClass
NameError: name 'x' is not defined

But the following would work:

>>> def myfunc():
...     x = 3
...     class MyClass(object):
...         y = x
...     return MyClass
... 
>>> myfunc().y
3

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