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Message-ID: <57180921.8020000@gmail.com>
Date: Wed, 20 Apr 2016 23:56:33 +0100
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Subject: Re: [PHP-DEV] [RFC:generics]
From: rowan.collins@gmail.com (Rowan Collins)

On 20/04/2016 22:02, Jesse Schalken wrote:
> "instanceof" implies to me that the right hand side is a class/interface
> name, since that's what's expected with the "instanceof" operator, rather
> than a type. If I can do "Foo<T instanceof array<Bar>>" I would expect to
> be able to do "if ($t instanceof array<Bar>) ...", but I can't.

I'm not quite sure what bounds would mean for anything other than 
classes or interfaces. A generic type that specified that its type 
parameter must be an int seems to me to be a non-generic type, because 
nothing other than an int "is int". How is "class Foo<T is int> { public 
function __construct(T $arg) { ... } }" different from just saying 
"class Foo { public function __construct(int $arg) { ... } }"? (Unless 
there were some kind of overloading, such that you could also have Foo<T 
is float>, but please let's not go there...)

Note that generic arrays are out of scope of the current RFC anyway, so 
it's possible that "$t instanceof array<Bar>" could be added at the same 
time as "Foo<T instanceof array<Bar>>", presumably with the meaning that 
"array<SubClassOfBar>" would be considered to pass the check.

Regards,

-- 
Rowan Collins
[IMSoP]