Question

While I was watching some code about singleton’s in python I decided to write my own.

This was my first code:

def singleton(cls):
    instance = False
    def constructor(*args,**kwargs):
        if not instance:
            instance = cls(*args,**kwargs)
        return instance
    return constructor

But when i tested it, the interpreter told me that ‘instance’ must be declared before being used on the if condition, finally I figured out to do it as follows:

def singleton(cls):
    cls._instance = False
    def constructor(*args,**kwargs):
        if not cls._instance:
            cls._instance = cls(*args,**kwargs)
        return cls._instance
    return constructor

and it worked as expected:

>>> @singleton
>>> class A: pass
>>> a=A()
>>> id(a)
33479456
>>> b=A()
>>> id(b)
33479456

Why the closure from the first example didn’t work.

Edit: the error was

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "singleton.py", line 4, in constructor
    if not instance:
UnboundLocalError: local variable 'instance' referenced before assignment

Answer 1

Your first closure didn’t work because inside your constructor function, you assigned to instance. That makes instance a local name inside constructor, and you accessed that local name before it was assigned to.

In Python 3, you can use nonlocal instance to declare the scope of instance. In Python 2, you can’t access that outer instance name as you want to.

Also, singletons are unusual in Python. Why not just instantiate your class once? Why try to trick Python into behaving differently than it does? Or create a factory function to produce the instance to use?