On 13 March 2012 21:40, Andrei Alexandrescu <seewebsiteforem...@erdani.org>wrote:
> On 3/13/12 1:20 PM, Manu wrote: > >> What value does it add over Kenji's change? Is this because Kenji's >> change is unable to perform direct to existing variables? >> > > Yes. > > > My understanding from early in the thread was that Kenji's change hides >> the returned tuple, and performs a convenient unpack. How can you >> perform a scatter if the tuple instance is no longer visible? >> > > If I understand you correctly, you just say fun().scatter(v1, v2, v3). Ah okay, I see. And you think that's more readable and intuitive than: (v1, v2, v3) = fun(); ?
