Improve metaclass conflict error message with clearer terminology and contextual information

Feature or enhancement

Proposal:

Improve Metaclass Conflict Error Message

Summary

Improve the error message raised during metaclass conflicts to specify the
base with incompatible metaclass.

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Motivation

Class inheritance has a well-established terminology across languages, but
Python's metaclass system introduces a distinct mechanism that lacks clear
linguistic conventions. This feature proposes the following
terminologies.

• A class is derived from its metaclass
• A class is based on its base classes
• A metaclass is any subclass of type

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Example

In the following code, MetaFoo and MetaBar are both subclasses of type
making them metaclasses according to the previous terminology suggestion.
Since neither is a subclass of the other, they are incompatible as metaclasses.
Foo derives from MetaFoo and Bar derives from MetaBar, but because
Bar is based on Foo; the conflicting metaclasses cause the TypeError
relating to metaclass conflict.

class MetaFoo(type): pass

class MetaBar(type): pass

class Foo(metaclass=MetaFoo): pass

class Bar(Foo, metaclass=MetaBar): pass  # raises

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Current Behaviour

The TypeError currently raised looks like this:

TypeError: metaclass conflict: the metaclass of a derived class must be a
(non-strict) subclass of the metaclasses of all its bases

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Proposed Behaviour

The improved error message would add contextual details and adopt the new
terminology: (Line breaks included for clarity).

Metaclass conflict while creating a new class!
- Declared metaclass: 'MetaBar'
- Incompatible base class: 'Foo'
- That base is derived from metaclass: 'MetaFoo'
All base classes must be based on classes derived from the same metaclass or a
subclass thereof.

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Notes

This change requires changes to the _PyType_CalculateMetaclass function only.
Including the name (Bar in the example) in the error message would require
changes to multiple functions across the codebase.

Has this already been discussed elsewhere?

This is a minor feature, which does not need previous discussion elsewhere

Links to previous discussion of this feature:

No response

Beginnen Sie mit der in der Issue genannten Funktion _PyType_CalculateMetaclass und reproduzieren Sie das Beispiel mit MetaFoo, MetaBar, Foo und Bar. Vergleichen Sie den resultierenden TypeError mit der vorgeschlagenen Formulierung, einschließlich der deklarierten Metaklasse, der inkompatiblen Basisklasse und ihrer Metaklasse; als erledigt gilt die Aufgabe, wenn der Konflikt diese Kontextinformationen unter Verwendung der angegebenen Terminologie meldet.