On Saturday, February 15, 2014 6:27:33 AM UTC+5:30, Chris Angelico wrote: > On Sat, Feb 15, 2014 at 8:43 AM, Marko Rauhamaa wrote: > > Unfortunately neither the "everything is a reference" model nor the > > "small/big" model help you predict the value of an "is" operator in the > > ambiguous cases.
> Can you give an example of an ambiguous case? Fundamentally, the 'is' > operator tells you whether its two operands are exactly the same > object, nothing more and nothing less, so I assume your "ambiguous > cases" are ones where it's possible for two things to be either the > same object or two indistinguishable ones. Fundamentally your definition above is circular: In effect the python expr "a is b" is the same as a is b. The only way to move ahead on that circularity is to 'leak-out' the under-belly of python's object-model. My own preference: No is operator; only id when we deliberately need to poke into the implementation. Of course I am in a miniscule minority I guess on that :-) -- https://mail.python.org/mailman/listinfo/python-list
