Yet another Python variable scope question

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 case.

• Rich: Why not just pass v to the ff() function?
• rsk: @Rich for example there are can be multiple nested functions

Answer 1

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()

• user395760: Note that the second falls to stackoverflow.com/questions/1132941/… (calling it a second time without providing v prints 2).

Answer 2

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.

• senderle: Perhaps def ff(v=v) would come closer to what the OP wants.