python closure local variables

问题

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.

回答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')

来源：https://stackoverflow.com/questions/11142993/python-closure-local-variables