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

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
scope

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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 这个 post. Additionally, please read PEP 3104 for more information on this subject.

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最佳答案

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

解决方法是使用一个可变对象并更新该对象的成员。