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.
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() 
v prints 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.
def ff(v=v) would come closer to what the OP wants.