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To: Josh Di Fabio <joshdifabio@gmail.com>
CC: Nikita Popov <nikita.ppv@gmail.com>, PHP internals
	<internals@lists.php.net>
Thread-Topic: [PHP-DEV] Re: [RFC] Arrow functions / short closures
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Date: Fri, 15 Mar 2019 21:44:26 +0000
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Subject: Re: [PHP-DEV] Re: [RFC] Arrow functions / short closures
From: theodorejb@outlook.com (Theodore Brown)

On Fri, March 15, 2019 at 4:45 AM Josh Di Fabio <joshdifabio@gmail.com> wro=
te:=0A=
=0A=
> I'd certainly be on board with the fn() syntax, but the backslash=0A=
> syntax has definitely grown on me. To me, all of the examples in=0A=
> Theodore's email are very readable and I find that the backslash makes=0A=
> it very easy to identify arrow functions when grokking.=0A=
>=0A=
> `array_filter($numbers, \$n =3D> $n % 2);`=0A=
> vs.=0A=
> `array_filter($numbers, fn($n) =3D> $n % 2);`=0A=
>=0A=
> When grokking these two pieces of code, I immediately see `\$n =3D> $n % =
> 2`=0A=
> as a single unit, whereas in the latter example I instinctively=0A=
> (and incorrectly) interpret "fn($n)" as an expression.=0A=
>=0A=
> When parens are required, the difference is obviously reduced, but I=0A=
> think I still prefer the backslash syntax since the LHS doesn't grok=0A=
> as a function call.=0A=
>=0A=
> `reduce($numbers, \($x, $y) =3D> $x + $y);`=0A=
> vs.=0A=
> `reduce($numbers, fn($x, $y) =3D> $x + $y);`=0A=
> =0A=
> That said, I'd personally take either syntax gladly.=0A=
=0A=
=0A=
You have a good point about `fn()` looking like an expression.=0A=
That said, since it would be a keyword IDEs will highlight it=0A=
differently which should help avoid confusion.=0A=
=0A=
Regarding the backslash syntax, I forgot to check how it looks with=0A=
by-reference passing and returning before I sent my email. Here are=0A=
those examples:=0A=
=0A=
```php=0A=
fn(&$x) =3D> $x;=0A=
fn&($x) =3D> $x;=0A=
=0A=
// vs.=0A=
=0A=
\(&$x) =3D> $x;=0A=
\&($x) =3D> $x;=0A=
=0A=
// unclear if passing or returning by reference=0A=
\&$x =3D> $x;=0A=
=0A=
// and worst of all...=0A=
fn&(&$x) =3D> $x;=0A=
=0A=
// vs.=0A=
\&(&$x) =3D> $x;=0A=
```=0A=
=0A=
I have to admit that the `fn` prefix is a little more readable for=0A=
these use cases (though I've never actually seen a real function=0A=
using by-reference passing and returning at the same time).=0A=
=0A=
-Theodore Brown=