python closure local variables

Question

In this answer a singleton decorator is demonstrated as such

def singleton(cls):
    instances = {}
    def getinstance():
        print len(instances)
        if cls not in instances:
            instances[cls] = cls()
        return instances[cls]
    return getinstance

but instances is 'local' to each class that is decorated, so I tried to be more efficient and use

def BAD_singleton(cls):
    instances = None
    def getinstance():
        if instances is None:
            instances = cls()
        return instances
    return getinstance

@BAD_singleton
class MyTest(object):
    def __init__(self):
        print 'test'

However, this gives an error

UnboundLocalError: local variable 'instances' referenced before assignment

when m = MyTest() is called

I think I understand which this should not work (as the assignment to instances will be local and be lost between calls), but I do not understand why I am getting this error.

@ephemient Thanks. That is a really useful looking page in general. — tacaswell
– tacaswell, Commented Jun 21, 2012 at 21:12

tacaswell · Accepted Answer · 2019-09-14 15:38:18Z

1

The reason for the error is python is cleverer than I am and identified that instances is made local by the assignment and does not go up-scope to find the assignment. As pointed out in the comments by @GeeTransit this is possible in python3 via nonlocal

def nonlocal_singleton(cls):
    instances = None
    def getinstance():
        nonlocal instances
        if instances is None:
            instances = cls()
        return instances
    return getinstance

@nonlocal_singleton
class MyTest(object):
    def __init__(self):
        print('test')

edited Sep 14, 2019 at 15:38

answered Jun 21, 2012 at 16:53

tacaswell

88.2k23 gold badges222 silver badges203 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

GeeTransit Over a year ago

It's been a while, but I'd like to point out that by adding a nonlocal instances, you can fix the error. It's raised because the function doesn't see a "declaration" or any sort before the first use of the variable. In this case, nonlocal instances would be similar to instances = some_value but some_value is the same as the one outside.

tacaswell Over a year ago

@GeeTransit Edited accordingly! Thanks!

Collectives™ on Stack Overflow

python closure local variables

1 Answer 1

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related