But on calling `type` in this way, you are passing the namespace as a
ready object, as the
3rdy parameter - what would `__prepare__` even do, besides print it's
been called?
For sometime (maybe Python 3.3, I forgot), some helper
callables where added to the `types` module to allow
one to have more freedom in imperatively created classes:
````python
In [20]: types.prepare_class?
Signature: types.prepare_class(name, bases=(), kwds=None)
Docstring:
Call the __prepare__ method of the appropriate metaclass.
Returns (metaclass, namespace, kwds) as a 3-tuple
*metaclass* is the appropriate metaclass
*namespace* is the prepared class namespace
*kwds* is an updated copy of the passed in kwds argument with any
'metaclass' entry removed. If no kwds argument is passed in, this will
be an empty dict.
File: ~/projetos/terminedia/env/lib64/python3.7/types.py
Type: function
In [21]: types.new_class?
Signature: types.new_class(name, bases=(), kwds=None, exec_body=None)
Docstring: Create a class object dynamically using the appropriate metaclass.
File: ~/projetos/terminedia/env/lib64/python3.7/types.py
Type: function
In [22]: types.resolve_bases?
Signature: types.resolve_bases(bases)
Docstring: Resolve MRO entries dynamically as specified by PEP 560.
File: ~/projetos/terminedia/env/lib64/python3.7/types.py
Type: function
```
On Mon, 1 Jun 2020 at 16:28, Ethan Furman <et...@stoneleaf.us> wrote:
>
> From stackoverflow [1]
>
> # metaprepare.py
> class Meta1(type):
> @classmethod
> def __prepare__(mcs, name, bases):
> print('call prepare')
> return {}
> def __new__(mcs, name, bases, parameters):
> return super().__new__(mcs, name, bases, parameters)
>
> class A(metaclass=Meta1):
> pass
>
> type('C', (A, ), {})
>
> output is:
>
> call prepare
>
> (just the once, not twice)
>
> The behavior of `type()` not calling `__prepare__()` has been constant since
> 3.3.
>
> Is it a bug?
>
>
> --
> ~Ethan~
>
>
> [1] https://stackoverflow.com/q/62128254/208880
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