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Message-ID: <560D8FCB.5080808@telia.com>
Date: Thu, 1 Oct 2015 21:55:55 +0200
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Subject: Re: [PHP-DEV] Arrow function expressions in PHP
From: bjorn.x.larsson@telia.com (=?UTF-8?Q?Bj=c3=b6rn_Larsson?=)

Den 2015-10-01 kl. 19:12, skrev Bishop Bettini:
> On Thu, Oct 1, 2015 at 12:28 PM, Anthony Ferrara <ircmaxell@gmail.com>
> wrote:
>
>> Nikita and all,
>>
>>> I don't think there was a dozen of different ideas, I could only find
>> those
>>> about `lambda(arg-list; use-list; expression)` and variations of it with
>>> different keywords and different return-type syntax.
>>> I do understand that this is quite subjective, but neither this syntax
>> nor
>>> `fn(arg-list; use-list) expression` look obvious and easily readable to
>> me.
>>
>> The problem is that semicolons are non-obvious, especially in a
>> context where commas are used (and traditionally used).
>>
>> Example, tell the difference quickly between
>>
>> fn($a, $b; $c) => $a + $b + $c;
>>
>> and
>>
>> fn($a; $b, $c) => $a + $b + $c.
>>
>> At a glance, they are identical. You have to actually look at each
>> item to realize that there's a difference. At least with use() there's
>> a syntactical separator there to draw your eye to the contextually
>> relevant information.
> True.  But a developer can mitigate with judicious white-space:
>
> fn($a, $b ; $c) => $a+$b+$c
>
> Or we can figure out some other such symbol. Worse casing no white space,
> brain storming:
>
> fn($a,$b:$c) => $a+$b+$c // not much better
> fn($a,$b!$c) => $a+$b+$c // better, but looks like not
> fn($a,$b&$c) => $a+$b+$c // lost in the noise, looks like bitwise
> fn($a,$b%$c) => $a+$b+$c // perl jibberish, looks modulo
> fn($a,$b--$c) => $a+$b+$c // multi-char, looks like decrement
> fn($a,$b::$c) => $a+$b+$c // maybe, kinda confusing
>
> // my favorite
> fn($a, $b @ $c) => $a + $b + $c;
>
Would it be: fn($a, $b @ $c, $d) => $a + $b + $c + $d;
with several parameters? And if one could solve the
parser problem it could be with typehint & a default
value either:
(int $a = 10, int $b @ $c, $d) => $a + $b + $c + $d;
or
(int $a = 10, int $b @ $c, $d) ==> $a + $b + $c + $d;
or
(int $a = 10, int $b @ $c, $d) ~> $a + $b + $c + $d;

Regards //Björn