Python Local Variable Scope Bug Within An Inner Function Definition

In Python, you can declare a function within another function and the scope of local variables in the outer function carry through to the inner function. However, you have to be careful with the re-assignment of the variable inside of the inner function due to some funkiness that may or may not be a bug (I say its a bug but who am I?)

Take the following example:

>>> def top_f(f):
>>>    return f()
>>>
>>> def outer_f(v=None):
>>>     print "v in outer = %s" % v
>>>     def inner_f():
>>>         print "v in inner = %s" % v
>>>     return top_f(inner_f)
>>>
>>> outer_f("see me?")
>>> v in outer = see me?
>>> v in inner = see me?

As expected the variable is visible within the inner function. Now look at the above example with one minor modification. We will re-assign the value of the variable in a nested fashion:

>>> def top_f(f):
>>>    return f()
>>>
>>> def outer_f(v=None):
>>>     print "v in outer = %s" % v
>>>     def inner_f():
>>>         v = v + " now"
>>>         print "v in inner = %s" % v
>>>     return top_f(inner_f)
>>>
>>> outer_f("see me?")
>>> v in outer = see me?
>>> Traceback (most recent call last):
...
>>> UnboundLocalError: local variable 'v' referenced before assignment

Whaaa?!?

In my opinion, the construct of v = v + " now" should be allowed as it is everywhere else in code and behave as expected. However, obviously something is amiss on the stack (or would that be heap?) – I care not to speculate as such low level memory thinking makes my head hurt and if I wanted to deal with such things I would code in C or worse, assembly!

This behaviour is present in v. 2.4 and 2.5. I have not tested in v. 2.6 or 3.0