RE: [PHP-DEV] [RFC] Spread Operator in Array Expression v0.2

Thanks Rowan. IMO it's very natural to try to apply operators to similar scenario, array construction in this case, and I'm completing the missing piece of jigsaw puzzle. Here is a MDN document for spread operator in JS: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax#Spread_in_array_literals and hope you find more useful examples. TL;DR: this RFC is not making something impossible to become possible, it's making something not so easy to become easy and efficient. -----Original Message----- From: Rowan Collins <rowan.collins@gmail.com> Sent: Friday, April 5, 2019 12:40 AM To: Derick Rethans <derick@php.net> Cc: CHU Zhaowei <me@jhdxr.com>; PHP internals <internals@lists.php.net> Subject: Re: [PHP-DEV] [RFC] Spread Operator in Array Expression v0.2 On Thu, 4 Apr 2019 at 17:14, Derick Rethans <derick@php.net> wrote: > Could you add to the RFC what the exact pain point is that this is > trying to address? It looks a little like this is just adding syntax > for the sake of it. > Not everything is about pain, some things are just about gain. ;) The link Levi shared about Dart included some interesting examples of where spreads are useful, some of which you can probably imagine happening in PHP: https://medium.com/dartlang/making-dart-a-better-language-for-ui-f1ccaf9f546c It also takes us a step closer to having a short-hand for iterator_to_array, in the shape of [...$iterator]. On its own, that's still pretty ugly, but it's not hard to come up with cases where it would be a lot nicer, like concatenating two iterators: // Before array_merge(iterator_to_array($iter1), iterator_to_array($iter2)) // Or to generalise to all iterables array_merge( is_array($iter1) ? $iter1 : iterator_to_array($iter1), is_array($iter2) ? $iter2 : iterator_to_array($iter2) ) // After (handles both cases) [ ...$iter1, ...$iter2 ] Granted, I can't point to a real-life example of that, but it shows that this isn't just new syntax for something that's already easy. Regards, -- Rowan Collins [IMSoP]