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Home/ Questions/Q 8150837
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T15:11:05+00:00 2026-06-06T15:11:05+00:00

In this answer a singleton decorator is demonstrated as such def singleton(cls): instances =

  • 0

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.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-06T15:11:06+00:00Added an answer on June 6, 2026 at 3:11 pm

    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')
    
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