Takuo Matsuoka <motogeom...@gmail.com> added the comment:
Thank you Ethan for reopening this issue and closing the other one.
Here is a description of a more specific issue, containing a more
reasonable example. I've changed the title of the issue to a more
appropriate one accordingly.
Context
-------
Some classes have the variable __name__ in their namespace __dict__ ,
and one may wish to create more such classes with varied values of
__name__ . Some of those could be created with a metaclass whose
__prepare__ returns a mapping having key "__name__", for which the
value is created depending on the arguments of __prepare__ and can be
updated or deleted in the body of the class to be created. (See C
below for a very silly example of such a metaclass.)
Problem
-------
The value of __name__ given by __prepare__ becomes not just that in
the class body, but automatically also the value of __module__
there. As far as I could see, this is not documented, and the
programmer might not notice __module__ was messed up. I think this
behaviour is unexpected and problematic at least unless a warning is
given on it in the document.
Also, the problem means we can't safely enjoy the ability of
__prepare__ of a metaclass to give a candidate for the value of
__name__ in __dict__ of the class without the trouble of fixing the
variable __module__ later at the top of the class body for every
instance of the metaclass (very annoying) somehow (or in __new__ or
__init__ of the metaclass if __module__ is not to be read in the class
body).
Example
-------
Here's a code which produces a problem.
```
# In this example, the metaclass C is intended to be a class of
# subclasses of:
B = type
class C(type(B)):
@classmethod
def __prepare__(cls, /, *args, **kwargs):
return dict(__name__ = cls._name(*args, **kwargs))
@classmethod
def _name(cls, /, *args, **kwargs):
# The actual value of __name__ doesn't matter much to the
# issue, so I make this function always return the same silly
# thing in this example.
return type.__dict__["__name__"]
class O(B, metaclass=C):
print(__module__ == __name__) # True
# Could update or delete __name__ here.
```
Consequently,
>>> O.__module__
<attribute '__name__' of 'type' objects>
Thanks.
P.S.
The argument mentioning "the scope outside" in my earlier post here
didn't make sense without specifying which scope. I still hope the
problem can be fixed.
----------
title: Wrong value assigned automatically to the variable __module__ in the
class body. -> The variable __module__ in the class body getting an undesirable
value from __prepare__ of the metaclass
_______________________________________
Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue47136>
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