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

[Joe Groff]

> 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).

-Joe