Yes, this is theoretically possible, but why is it useful? (I am genuinely
curious.)
If the intention is to allow B to be a user-defined extension point for T,
this goes back to the core team's arguments (in the thread about optional
protocol requirements) about why having checking for conformance to some
requirement at the use site is a suboptimal idea.
If the intention is to make the type system as expressive as possible, the
core team has already ruled out a number of features (user-defined variance
on generics, generic protocols) because they don't believe in their general
applicability.
Austin
On Sat, May 14, 2016 at 11:13 AM, Vladimir.S <sva...@gmail.com> wrote:
> FWIW, yes, protocols available for struct are known at compile-time, but
> could be unknown at the *moment of writing* the code.
>
> What I mean:
>
> Step 1. I write source code:
>
> protocol A {}
> protocol B {}
> struct S:A {}
>
> func f(a: A) {
> if a is struct<S,B> {...} // I expect that S could be conformed to B
> }
>
> Step 2. I give my code to someone, who can do somewhere in his project:
>
> extension S:B{..}
>
>
>
>
> On 14.05.2016 7:06, Austin Zheng via swift-evolution wrote:
>
>> 1. struct<SomeConcreteStruct, Protocol1, Protocol2>. In this case the
>> struct<...> representation is unnecessary; the protocols that are
>> available
>> to the user are known at compile-time, and structs can't have subtypes
>> that
>> conform to additional protocols like classes can. There is an example
>> marked "func boo(value: struct<SomeStruct>) /* equivalent to */ func
>> boo(value: SomeStruct)"; my question is why having more than two ways to
>> express the same idea makes the language better, easier to use, etc.
>>
>>
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