Yet another Python variable scope question

I have the following code:

>>> def f(v=1):
...     def ff():
...             print v
...             v = 2
...     ff()
...
>>> f()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 5, in f
  File "<stdin>", line 3, in ff
UnboundLocalError: local variable 'v' referenced before assignment

I do understand why this message occurs (Python variable scope question), but how can I work with v variable in this case? global v doesn't work in this开发者_Go百科 case.

In Python 3.x, you can use nonlocal:

def f(v=1):
    def ff():
        nonlocal v
        print(v)
        v = 2
    ff()

In Python 2.x, there is no easy solution. A hack is to make v a list:

def f(v=None):
    if v is None:
        v = [1]
    def ff():
        print v[0]
        v[0] = 2
    ff()

You haven't passed v to the inner function ff. It creates its own scope when you declare it. This should work in python 2.x:

def f(v=1):
  def ff(v=1):
    print v
    v = 2
  ff(v)

But the assignment call to v = 2 would not be persistent in other calls to the function.