https://github.com/python/cpython/commit/d1eefcb1a56a3cbdff176553ae892a24cc47488b commit: d1eefcb1a56a3cbdff176553ae892a24cc47488b branch: 3.13 author: Miss Islington (bot) <[email protected]> committer: encukou <[email protected]> date: 2026-07-29T15:33:08+02:00 summary:
[3.13] gh-62534: Document that three-argument type() does not call __prepare__ (GH-154028) (GH-154164) 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. (cherry picked from commit 5625b187ab528e10dcaa21726a7c95b110d97190) Co-authored-by: soreavis <[email protected]> files: M Doc/library/functions.rst diff --git a/Doc/library/functions.rst b/Doc/library/functions.rst index e7655449154a18..a42d0f338577f5 100644 --- a/Doc/library/functions.rst +++ b/Doc/library/functions.rst @@ -2091,6 +2091,11 @@ 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 _______________________________________________ Python-checkins mailing list -- [email protected] To unsubscribe send an email to [email protected] https://mail.python.org/mailman3//lists/python-checkins.python.org Member address: [email protected]
