[cessor]

Members in Python are never private. Variables that are supposed to be used internally only are marked with and underscore, but that just conveys intent and isn't enforced by the interpreter/runtime. But you can emulate private data like this:

    >>> def a(u, v):
    ...     def b():
    ...         return u + v
    ...     return b
    ...
    >>> b = a(1,2)
    >>> b()
    3

Like this, there is no way to access u & v from b.

[moonchild]

Looks like classes can't be closures, though:

  >>> def f():
  ...     x = 5
  ...     class Blub:
  ...             def incx(self):
  ...                     x += 1
  ...             def getx(self):
  ...                     return x
  ...     return Blub()
  ... 
  >>> j = f()
  >>> j.incx()
  Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
    File "<stdin>", line 5, in incx
  UnboundLocalError: local variable 'x' referenced before assignment

[cessor]

When using it like this, the interpreter opens a new namespace where the variables are bound. The x is declared in the outer scope, so it can't find it. However, you can ask python to keep searching vor the name `x` in the closest outer scope, that is what the `nonlocal` statement is for:

    def f():
        x = 5
        class Blub:
            def incx(self):
                nonlocal x
                x += 1

            def getx(self):
                return x

        return Blub()

    j = f()
    j.incx()
    print(j.getx())

This will print 6, as expected.