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Date: Tue, 4 Nov 2014 16:28:56 +0000
Cc: Stas Malyshev <smalyshev@sugarcrm.com>,
 Robert Stoll <php@tutteli.ch>,
 PHP Internals <internals@lists.php.net>
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To: Chris Wright <cw@daverandom.com>
Subject: Re: [PHP-DEV] Types on the right or on the left
From: ajf@ajf.me (Andrea Faulds)


> On 4 Nov 2014, at 11:50, Chris Wright <cw@daverandom.com> wrote:
>=20
> On 4 November 2014 10:32, Andrea Faulds <ajf@ajf.me> wrote:
>=20
> > I thought it was inconsistent, but after discussions on =
StackOverflow, I don't think it actually is. > Return types describe the =
return type of a function, not the type of a function. So there's no =
reason > they have to go before the function name like parameter types =
do.
>=20
> So, in a sentence:
>=20
> In PHP, input types go on the left and output types go on the right.
>=20
> This sounds like a convention that could be established and followed, =
and no-one could reasonably call it inconsistent if it was adhered to.
>=20

That=E2=80=99s not quite what I meant, but I=E2=80=99d also agree with =
that convention.

For parameters, the type is of the thing itself:

    function foo(Foo $foo, Bar $bar);

Here, $foo is of the type Foo, and $bar is of the type Bar.

But say if we were to add return types like this:

    function FooBar qux();

The type of qux isn=E2=80=99t a FooBar, it=E2=80=99s a function (more =
accurately, an argument-less, return-anything function). FooBar isn=E2=80=99=
t what qux is, it=E2=80=99s not qux=E2=80=99s type, merely what it =
returns. In a sense, I=E2=80=99d say there=E2=80=99s already a type =
here: `function`.

Compare it to this:

    function qux(): FooBar;

FooBar is what qux returns, `function` is what qux is.

If we were to hypothetically add typed properties with the same syntax =
Hack has, they=E2=80=99d look like this:

    public Foo $foo;

This would be consistent with parameters. And I think it goes along well =
with return types at the end:

    public Foo $foo;
    public function foobar(): FooBar;

Here, it=E2=80=99s clear the second item is a function, and the previous =
item is a property of type Foo.

=E2=80=A6but hey, I could be wrong. Does this make sense to anyone?

--
Andrea Faulds
http://ajf.me/