Why nested functions can access variables from outer functions, but are not allowed to modify them

Question

In the 2nd case below, Python tries to look for a local variable. When it doesn't find one, why can't it look in the outer scope like it does for the 1st case?

This looks for x in the local scope, then outer scope:

def f1():
    x = 5
    def f2():
         print x

This gives local variable 'x' referenced before assignment error:

def f1():
    x = 5
    def f2():
        x+=1

I am not allowed to modify the signature of function f2() so I can not pass and return values of x. However, I do need a way to modify x. Is there a way to explicitly tell Python to look for a variable name in the outer scope (something similar to the global keyword)?

Python version: 2.7

python 3 provides `nonlocal` for this purpose, but I think it's worth asking why you need to do this. This is a little safer than using `global`, but still doesn't feel right. — mgilson, Aug 16 '12 at 12:52
@mgilson there are many good reasons to do this, see e.g. the rationale section in pep-3104. — thebjorn, Aug 16 '12 at 12:58
@Dhara sorry, don't know why the comment got posted. It was fairly obvious that you're on 2.7. — soulcheck, Aug 16 '12 at 13:06

score 187 · Answer 1 · edited May 23 '17 at 12:02

187

In Python 3.x this is possible:

def f1():
        x = 5
        def f2():
                nonlocal x
                x+=1
        return f2

The problem and a solution to it, for Python 2.x as well, are given in this post. Additionally, please read PEP 3104 for more information on this subject.

edited May 23 '17 at 12:02

Community

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answered Aug 16 '12 at 12:49

hochl

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I just looked this up. `nonlocal` is the statement to use here. – mgilson Aug 16 '12 at 12:50
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@user1320237 global does not work instead of nonlocal -- that was precisely my question! – Dhara Aug 16 '12 at 18:11

score 79 · Accepted Answer · answered Aug 16 '12 at 12:50

79

def f1():
    x = { 'value': 5 }
    def f2():
        x['value'] += 1

Workaround is to use a mutable object and update members of that object. Name binding is tricky in Python, sometimes.

answered Aug 16 '12 at 12:50

Tom Whittock

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This works, although I ended up using a list with a single element instead of a dictionary. Thanks! – Dhara Aug 17 '12 at 07:21
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I think it is more correct to use `nonlocal` or `global` – bgusach Dec 03 '13 at 08:39
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Global is not a good idea as it will pollute the global namespace. Nonlocal is a python 3 feature, and this question specified python 2.7. – Tom Whittock Dec 04 '13 at 11:08
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*global* doesn't work in py2.7 for question purpose, however `x['value']` is harder to write on parent scope than just `value`, so consider passing value to nested function, and then just returning `value` from nested scope to parent scope – Sławomir Lenart Jul 18 '18 at 15:59

Why nested functions can access variables from outer functions, but are not allowed to modify them

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