> So for a non host object o, > o.x = o.x - 1 > should yield a number regardless of the details of any setter for x on > o, assuming o is not frozen and said setter returns normally.
The assignment operator is defined in the language fairly clearly in a way that means that the behaviour is unaffected by the type of object being assigned to -- eg. being a host object or not is (mercifully) not relevant :D --Oliver _______________________________________________ es-discuss mailing list [email protected] https://mail.mozilla.org/listinfo/es-discuss
