Raymond Hettinger wrote:
If an external non-cooperative class needs to be used, then it should be wrapped in a class that makes an explicit __init__ call to the external class and then calls super().__init__() to continue the forwarding.
I don't think it's as simple as that. Isn't that super() call going to call the __init__() method that you just explicitly called *again*? Seems like you would at least need to use super(BaseClass)... to skip the one you just called. But it's not immediately obvious to me that this won't ever skip other classes that you *do* want to call. -- Greg _______________________________________________ Python-Dev mailing list Python-Dev@python.org http://mail.python.org/mailman/listinfo/python-dev Unsubscribe: http://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
