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
--
Marco
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