gh-62534: Document that three-argument type() does not call __prepare__#154028
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…repare__ 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.
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Thanks for fixing this long-standing omission! |
encukou
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needs backport to 3.13
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GH-154164 is a backport of this pull request to the 3.13 branch. |
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GH-154165 is a backport of this pull request to the 3.15 branch. |
bedevere-app
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needs backport to 3.15
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GH-154166 is a backport of this pull request to the 3.14 branch. |
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The three-argument form of
type()skips the metaclass__prepare__method —__prepare__is called by theclassstatement machinery, not by the metaclass call itself — so a class created withtype('C', (Parent,), {})silently misses anything a base's metaclass prepares. The entry currently calls the form "essentially a dynamic form of the class statement" and documents the keyword-argument metaclass machinery, but never mentions this difference.Add the note the issue thread settled on: state that the three-argument form does not call
__prepare__, and point totypes.new_class()for dynamic class creation with the appropriate metaclass. Raising instead was considered and rejected in the thread, since a metaclass's own__new__legitimately callstype(name, bases, ns).Verified at tip: a subclass created via the
classstatement ortypes.new_class()gets the prepared namespace; the same bases through three-argumenttype()do not.