[swift-evolution] [swift-evolution-announce] [Review] SE-0096: Converting dynamicType from a property to an operator

[Joe Groff]

> On May 25, 2016, at 12:58 PM, Dave Abrahams <dabrahams at apple.com> wrote:
> 
> 
> on Wed May 25 2016, Joe Groff <jgroff-AT-apple.com> wrote:
> 
>>> On May 25, 2016, at 12:28 PM, Dave Abrahams <dabrahams at apple.com> wrote:
>>> 
>>> 
>>> on Wed May 25 2016, Joe Groff <jgroff-AT-apple.com> wrote:
>>> 
>> 
>>>>> On May 25, 2016, at 11:42 AM, Erica Sadun <erica at ericasadun.com> wrote:
>>>>> 
>>>>> 
>>>>>> On May 25, 2016, at 12:26 PM, Dave Abrahams via swift-evolution <swift-evolution at swift.org> wrote:
>>>>>> I don't understand why the proposal says we can't implement this in the
>>>> 
>>>>>> library today.  
>>>>>> 
>>>>>> $ swift
>>>>>> Welcome to Apple Swift version 2.2 (swiftlang-703.0.18.8 clang-703.0.30). Type :help for assistance.
>>>>>> 1> func dynamicType_<T>(_ x: T) -> T.Type { return x.dynamicType }
>>>>>> 2> dynamicType_(42)
>>>>>> $R0: Int.Type = Int
>>>>>> 3> class B {}
>>>>>> 4. class C : B {}
>>>>>> 5. dynamicType_(C() as B)
>>>>>> $R1: B.Type = __lldb_expr_5.C
>>>>>> 6>
>>>> 
>>>> Now try it with a protocol type, or Any:
>>>> 
>>>> (swift) var x: Any = 1738
>>>> // x : Any = 1738
>>>> (swift) dynamicType_(x)
>>>> // r0 : Any.Protocol = protocol<>
>>>> 
>>>> `dynamicType` is really two operations: For normal concrete types, it
>>>> produces concrete metatypes, and for existentials, it produces
>>>> existential metatypes. There's no way to express the latter for an
>>>> arbitrary unknown protocol type in the language today.
>>> 
>>> Can't we detect in the runtime library that we've got an existential and
>>> do the right thing?
>> 
>> Not within the constraints of the type system. P.Protocol and P.Type
>> are different types, and the former isn't a model of the latter (since
>> P has no methods of its own so can't satisfy P's static requirements).
> 
> I mean at runtime, in C++ code.

When you substitute 'T = P' into 'T.Type', which specifies a concrete metatype, you get 'P.Protocol'. We can't return P.Type without breaking the type of the function.

-Joe