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Date: Thu, 17 Jun 2021 09:54:26 +0200
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Subject: Re: [PHP-DEV] [Vote] Partial Function Application
From: come.chilliet@fusiondirectory.org (=?UTF-8?B?Q8O0bWU=?= Chilliet)

Le Wed, 16 Jun 2021 11:16:28 -0500,
"Larry Garfield" <larry@garfieldtech.com> a =C3=A9crit :

> Hi folks.  The vote for the Partial Function Application RFC is now open,=
 and
> will run until 30 June.
>=20
> https://wiki.php.net/rfc/partial_function_application

I do not understand how this ... placeholder works, it feels inconsistent.

=46rom the examples:

> $c =3D stuff(...);
> $c =3D fn(int $i, string $s, float $f, Point $p, int $m =3D 0)
>   =3D> stuff($i, $s, $f, $p, $m);

> $c =3D stuff(1, 'hi', 3.4, $point, 5, ...);
> $c =3D fn(...$args) =3D> stuff(1, 'hi', 3.4, $point, 5, ...$args);

Why is there an additional variadic parameter in this one?

If I remove the 5:
> $c =3D stuff(1, 'hi', 3.4, $point, ...);
I get this right?:
> $c =3D fn(int $m =3D 0) =3D> stuff(1, 'hi', 3.4, $point, $m);

Why the ... does not produce 0 parameters when there are none left?

Also, in the second set of examples:
> function things(int $i, float $f, Point ...$points) { ... }
=20
> // Ex 13
> $c =3D things(...);
> $c =3D fn(int $i, float $f, ...$args) =3D> things(...[$i, $f, ...$args]);
=20
> // Ex 14
> $c =3D things(1, 3.14, ...);
> $c =3D fn(...$args) =3D> things(...[1, 3.14, ...$args]);

What happens to the typing of the variadic parameter here? Why is it remove=
d?

It would feel natural that the ... means "copy the rest of the parameters f=
rom
 signature". Here it seems it sometimes mean that, and sometimes mean "acce=
pt an
 additional variadic parameter and pass it along".

C=C3=B4me