Comment by 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.