python: How do I capture a variable declared in a non global ancestral outer scope?

Question

Given:

def f():
    x = 0
    def g():
        h()
    def h():
        x += 1
        print(x)
    g()

>>> f()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 8, in f
  File "<stdin>", line 4, in g
  File "<stdin>", line 6, in h
UnboundLocalError: local variable 'x' referenced before assignment
>>>

How can I make h see the x variable?

Thanks.

EDIT

Should have mentioned it earlier, I am using Python 2.7.3

Answer 1

You can make x a function attribute :

def f():
    f.x = 0
    def g():
        h()
    def h():
        f.x += 1
        print(f.x)
    g()

Also, as of Python 3, you can use nonlocal keyword.

Answer 2

If you're using Python 3, you use the nonlocal keyword. Put nonlocal x at the beginning of function h . If you're using Python 2.x, a workaround is making x a list with one element, so you can modify it:

def f():
    x = [0]
    def g():
        h()
    def h():
        x[0] += 1
        print x[0]
    g()

f()

Answer 3

In Python 3 just use nonlocal :

def f():
    x = 0
    def g():
        h()
    def h():
        nonlocal x
        x += 1
        print(x)
    g()
f()

Answer 4

Can't we put x as function arguments as workaround

def f():
    x = 0
    def g():
        h(x)
    def h(x):
        x += 1
        print(x)
    g()

f() 

Answer 5

Easiest is to use a dict or empty class, eg:

class Empty:
    x = 0

def f():
    closure1 = dict(x=0)
    closure2 = Empty()
    def g(): h(x)
    def h(x):
        closure1["x"] += 1
        closure2.x += 1
    g()
    print closure1["x"], closure2.x

Although many good solutions were already provided, they have corner cases:

• nonlocal, per Ashwini, is Python 3.x only
• function attribute, per ovgolovin, will fail is f is redefined and later called by reference