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

issue で指定されている _PyType_CalculateMetaclass 関数から始め、MetaFoo、MetaBar、Foo、Bar の例を再現します。結果の TypeError を提案された文言と比較し、宣言されたメタクラス、互換性のない基底クラス、およびそのメタクラスが含まれていることを確認します。完了の条件は、競合が指定された用語を使ってそのコンテキスト情報を報告することです。