On Thu, 29 Sep 2005, Pierre Barbier de Reuille wrote: > a discussion began on python-dev about this. It began by a bug report, > but is shifted and it now belongs to this discussion group. > > The problem I find with augmented assignment is it's too complex, it's > badly explained, it's error-prone. And most of all, I don't see any > use-case for it ! > > The most common error is to consider that : > > a += b <==> a.__iadd__(b) > > when the truth is : > > a += b <==> a = a.__iadd__(b) > > which can be very confusing, as the two "a" are not necessarily the > same.
Indeed. I certainly didn't realise that was how it worked. > So, what I would suggest is to drop the user-defined augmented > assignment and to ensure this equivalence : > > a X= b <=> a = a X b > > with 'X' begin one of the operators. That seems quite an odd move. Your proposal would lead to even more surprising behaviour; consider this: a = [1, 2, 3] b = a a += [4, 5, 6] print b At present, this prints [1, 2, 3, 4, 5, 6]; if we were to follow your suggestion, it would be [1, 2, 3]. So, -1, i'm afraid. I think the right solution here is staring us in the face: if everyone seems to think that: a += b <==> a.__iadd__(b) Then why not make it so that: a += b <==> a.__iadd__(b) Principle of Least Surprise and all that. Since not everything that should support += is mutable (integers, for example), how about saying that if the recipient of a += doesn't have an __iadd__ method, execution falls back to: a = a + b I say 'a + b', because that means we invoke __add__ and __radd__ appropriately; indeed, the __add__ vs __radd__ thing is a precedent for this sort of fallback. Doesn't that leave everyone happy? tom -- I'm not quite sure how that works but I like it ... -- http://mail.python.org/mailman/listinfo/python-list
