On Sun, Mar 10, 2019 at 3:16 AM Jonathan Fine <jfine2...@gmail.com> wrote: > > Anders Hovmöller wrote: > > > I don't understand what you mean. Can you provide examples that show the > > state of the dicts before and after and what the syntax would be the > > equivalent of in current python? > > If a.__radd__ exists, then > a += b > is equivalent to > a = a.__radd__(b) > > Similarly, if a.__iat_update__ exists then > a @update= b > would be equivalent to > a = a.__iat_update__(b) > > Here's an implementation > def __iat_update__(self, other): > self.update(other) > return self > > Thus, 'b' would be unchanged, and 'a' would be the same dictionary as > before, but updated with 'b'.
With something this long, how is it better from just writing: a = a.update_with(b) ? What's the point of an operator, especially if - by your own statement - it will backward-incompatibly change the language grammar (in ways that I've yet to understand, since you haven't really been clear on that)? ChrisA _______________________________________________ Python-ideas mailing list Python-ideas@python.org https://mail.python.org/mailman/listinfo/python-ideas Code of Conduct: http://python.org/psf/codeofconduct/
