https://github.com/python/cpython/commit/5625b187ab528e10dcaa21726a7c95b110d97190
commit: 5625b187ab528e10dcaa21726a7c95b110d97190
branch: main
author: soreavis <[email protected]>
committer: encukou <[email protected]>
date: 2026-07-19T17:31:59+02:00
summary:
gh-62534: Document that three-argument type() does not call __prepare__
(GH-154028)
The three-argument form of type() skips the metaclass __prepare__
method, which is called by the class statement machinery rather than
by the metaclass call itself. Say so in the type() entry and point to
types.new_class() for dynamic class creation with the appropriate
metaclass, as directed in the issue thread.
files:
M Doc/library/functions.rst
diff --git a/Doc/library/functions.rst b/Doc/library/functions.rst
index 8dbebf7048b583..f45ab397e93693 100644
--- a/Doc/library/functions.rst
+++ b/Doc/library/functions.rst
@@ -2179,6 +2179,11 @@ are always available. They are listed here in
alphabetical order.
in the same way that keywords in a class
definition (besides *metaclass*) would.
+ Unlike a :keyword:`class` statement, the three argument form does not
+ call the metaclass ``__prepare__`` method (see :ref:`prepare`). Use
+ :func:`types.new_class` to dynamically create a class using the
+ appropriate metaclass.
+
See also :ref:`class-customization`.
.. versionchanged:: 3.6
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