Dennis Sweeney <sweeney.dennis...@gmail.com> added the comment:
typing.cast doesn't actually do anything, it only exists as a hint for type-checkers. As William noted, using the 3-argument type(...) as you showed will only return a type, not a mcs. I think you may want super().__new__(mcs, name, bases, namespace), which will return an instance of mcs. You could also write type.__new__(mcs, name, bases, namespace), but that would make multiple inheritance harder should you ever want to do that. Another note: x(*args) translates to type(x).__call__(x, *args), so whether or not I am callable depends not on whether I have a __call__ attribute, but rather on whether my type has a __call__ attribute See also: https://stackoverflow.com/q/6760685/11461120 ---------- nosy: +Dennis Sweeney _______________________________________ Python tracker <rep...@bugs.python.org> <https://bugs.python.org/issue43685> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
