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Re: Overriding of the type.__call__() method in a metaclass

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Marco Buttu wrote:

> Hi all, I have a question about class creation and the __call__ method.
> I have the following metaclass:
> 
>  >>> class FooMeta(type):
> ...     def __call__(metacls, name, bases, namespace):
> ...         print("FooMeta.__call__()")
> 
> 
>  From what I undestood, at the end of the class statement happens
> something like this:
> 
>  >>> def __call__(metacls, name, bases, namespace):
> ...     print("FooMeta.__call__()")
> ...
>  >>> FooMeta = type('FooMeta', (type,), {'__call__': __call__})
> 
> The call to the metaclass type causes the call to type.__call__(), so
> that's happened is:
> 
>  >>> FooMeta = type.__call__(type, 'FooMeta', (type,), {'__call__':
> __call__})
> 
> Now I expected the output `FooMeta.__call__()` from the following Foo
> class creation:
> 
>  >>> class Foo(metaclass=FooMeta):
> ...     pass
> 
> because I thought at the end of the class Foo suite this should have
> been happened:
> 
>  >>> Foo = FooMeta.__call__(FooMeta, 'Foo', (), {})
> FooMeta.__call__()
> 
> but I thought wrong:
> 
>  >>> class FooMeta(type):
> ...     def __call__(metacls, name, bases, namespace):
> ...         print("FooMeta.__call__()")
> ...
>  >>> class Foo(metaclass=FooMeta):
> ...     pass
> ...
>  >>>
> 
> How come? Is it because the first argument of metaclass.__call__() is
> always type or I am thinking something wrong?
> Thanks in advance, Marco

Forget about metaclasses for the moment and ask yourself what happens when a 
regular class

class A:
   def __init__(...): ...
   def __call__(...): ...

is "called":

a = A(...) # invokes __init__()
a(...) # invokes __call__()


The metaclass is just the class of a class, i. e. the Foo object is an 
instance of FooMeta, so making Foo invokes (__new__() and) __init__(), and 
calling Foo invokes FooMeta.__call__():

>>> class FooMeta(type):
...     def __call__(self, *args): print("__call__%r" % (args,))
... 
>>> class Foo(metaclass=FooMeta): pass
... 
>>> Foo()
__call__()

If you follow that logic you can easily see that for FooMeta to invoke your 
custom __call__() method you'd have to define it in FooMeta's metaclass:

>>> class FooMetaMeta(type):
...     def __call__(*args): print(args)
... 
>>> class FooMeta(metaclass=FooMetaMeta):
...     pass
... 
>>> class Foo(metaclass=FooMeta):
...     pass
... 
(<class '__main__.FooMeta'>, 'Foo', (), {'__module__': '__main__', 
'__qualname__': 'Foo'})
>>> Foo is None                                                                                                                                                                                                   
True                                                                                                                                                                                                              

So, yes, it's turtles all the way down...

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Thread

Overriding of the type.__call__() method in a metaclass Marco Buttu <marco.buttu@gmail.com> - 2013-10-06 20:17 +0200
  Re: Overriding of the type.__call__() method in a metaclass Peter Otten <__peter__@web.de> - 2013-10-06 21:04 +0200
  Re: Overriding of the type.__call__() method in a metaclass Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2013-10-07 02:27 +0000
    Re: Overriding of the type.__call__() method in a metaclass Marco Buttu <marco.buttu@gmail.com> - 2013-10-07 07:22 +0200

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