Python and closed variables

Question

Have a look at this code:

def closure():
    value = False

    def method_1():
        value = True

    def method_2():
        print 'value is:', value

    method_1()
    method_2()

closure()

I would expect it to print 'Value is: True' but it doesn't. Why is this and what is the solution?

Answer 1

This happens because method_1 gets its own local scope where it can declare variables. Python sees value = True and thinks you're creating a new variable named value, local to method_1.

The reason Python does this is to avoid polluting the outer scope's locals with variables from an inner function. (You wouldn't want assignments in regular, module-level functions to result in global variables being created!)

If you don't assign to value, then Python searches the outer scopes looking for the variable (so reading the variable works as expected, as demonstrated by your method_2).

One way to get around this is by using a mutable object instead of assigment:

result = { 'value': False }

def method_1():
    result['value'] = True

In Python 3, the nonlocal statement (see also docs) was added for exactly this scenario:

def method_1():
    nonlocal value
    value = True    # Works as expected -- assigns to `value` from outer scope

Answer 2

In method_1, Python assumes (quite sensibly!) that value is a local variable. Whenever you assign to a variable name inside a function, it is assumed that that variable name is a new local variable. If you want it to be global, then you have to declare it as global, and if you want it to be "nonlocal", in Python 3, you can declare it nonlocal, but in Python 2, you have to do something uglier: store the value in a container. That avoids having to reassign the variable name, and so avoids the scoping ambiguity.

def method_1_global():
    global value
    value = True

def method_1_nonlocal_P3():
    nonlocal value
    value = True

value = [False]
def method_1_nonlocal_P2():
    value[0] = True