need to understand the flow of __init__, __new__ and __call__


Question

class Singleton(type):
    def __init__(self, *args, **kwargs):
        print 'calling __init__ of Singleton class', self
        print 'args: ', args
        print 'kwargs: ', kwargs
        super(Singleton, self).__init__(*args, **kwargs)
        self.__instance = None
    def __call__(self, *args, **kwargs):
        print 'running __call__ of Singleton', self
        print 'args: ', args
        print 'kwargs: ', kwargs, '\n\n'
        if self.__instance is None:
            self.__instance = super(Singleton, self).__call__(*args, **kwargs)
        return self.__instance

class A(object):
    __metaclass__ = Singleton
    def __init__(self,a):
        print 'in __init__ of A:  ', self
        self.a = a
        print 'self.a: ', self.a

a=A(10)
b=A(20)

I copied this code from Ben's answer to the question Python's use of __new__ and __init__? and modified it a little. But, I am not aware of the flow. Although I understand from a higher level what this code is intended to do. But, internally how it works, I am not quite sure of.

I get the following output when running this code:-

calling __init__ of Singleton class <class '__main__.A'>
args:  ('A', (<type 'object'>,), {'__module__': '__main__', '__metaclass__': <class '__main__.Singleton'>, '__init__': <function __init__ at 0x01F9F7B0>})
kwargs:  {}
running __call__ of Singleton <class '__main__.A'>
args:  (10,)
kwargs:  {}


in __init__ of A:   <__main__.A object at 0x01FA7A10>
self.a:  10
running __call__ of Singleton <class '__main__.A'>
args:  (20,)
kwargs:  {}

I cant understand how the args and kwargs for __init__ and __call__ become different. While using metaclasses, this link (What is a metaclass in Python?) has explained how to use __new__ and a function as a metaclass. But, I do not understand how __call__ is being used.

Can somebody explain the flow? By this, I mean, the precedence in which __new__, __call__, __init__ are called and who calls them?

1
8
5/23/2017 12:10:28 PM

Accepted Answer

Your code doesn't include any __new__, so little can be said about it.

But you create a metaclass which is instantiated at the time class A is created. In other words, the class A is an object itself and as such an instance of its metaclass Singleton.

So let's look what happens:

After A's environment is finished (its methods exist, its dict exists as well, ...), the class gets created as an instance of the metaclass. Essentially, the call is

A = Singleton('A', (object,), <the dict>)

where <the dict> is the dict containing the class's namespace (here: __module__, __metaclass__ and __init__).

On this call to Singleton, calling super(Singleton, self).__call__(*args, **kwargs) results in calling the __new__ method which returns a new instance, on which .__init__ is called afterwards.

That's why this happens:

calling __init__ of Singleton class <class '__main__.A'>
args:  ('A', (<type 'object'>,), {'__module__': '__main__', '__metaclass__': <class '__main__.Singleton'>, '__init__': <function __init__ at 0x01F9F7B0>})
kwargs:  {}

After A is constructed, you use it by instantiating it:

a = A(10)

This calls A. A is an instance of Singleton, so Singleton.__call__ is invoked – with the effect you see:

running __call__ of Singleton <class '__main__.A'>
args:  (10,)
kwargs:  {}

Singleton.__call__ calls type.__call__, this calls A.__new__ and A.__init__:

in __init__ of A:   <__main__.A object at 0x01FA7A10>
self.a:  10

Then you do

b = A(20)

which calls Singleton.__call__:

running __call__ of Singleton <class '__main__.A'>
args:  (20,)
kwargs:  {}

Here the super call is suppressed and the old object is returned.

8
10/19/2012 5:59:18 PM

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