[issue18334] type(name, bases, dict) does not call metaclass' __prepare__ attribute

[Nick Coghlan]

Nick Coghlan added the comment:

Unfortunately, it's not that simple, as calling type(name, bases, namespace) is *exactly* what a subclass will do as part of creating the type object.

>From inside the type implementation, we can't tell the difference between "properly called from the child type" and "improperly called without preparing the namespace first".

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Python tracker <report at bugs.python.org>
<http://bugs.python.org/issue18334>
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