[swift-evolution] [Manifesto] Completing Generics

[Thorsten Seitz]

> Am 10.03.2016 um 05:35 schrieb Howard Lovatt via swift-evolution <swift-evolution at swift.org>:
> 
> I am sure you know the answer for Swift; but in other languages it doesn't, e.g. Java. 
> 
> With Java semantics:
> 
> func foo<Input, Output>(input: Input) -> Output
> 
> Would be the same as:
> 
> func foo(input: Any) -> Any
> 
> Because the only constraint on both Input and Output is Any. Since the Java compiler erases generics, the two would literally be the same to the JVM!

To the JVM at runtime, but not to the typechecker at compile time. There it works just as in Swift.

-Thorsten 


> Since Swift doesn't allow overloading on return type and therefore as it stands `func foo<Input, Output>(input: Input) -> Output` cannot be implemented, it would be nice if the compiler flagged this as an error and suggested that you meant `func foo(input: Any) -> Any`.
> 
>   -- Howard.
> 
>> On 10 March 2016 at 13:07, Joe Groff <jgroff at apple.com> wrote:
>> 
>> > On Mar 9, 2016, at 6:02 PM, Howard Lovatt <howard.lovatt at gmail.com> wrote:
>> >
>> > Wow, I would never have guessed that syntax :)
>> >
>> > It makes no sense to me to interpret a generic constraint as meaning all instead of any. How could anything either accept or return all possible implementations of something simultaneously, surely it only ever accepts or returns one of all the possible implementations at a time.
>> 
>> A type variable in angle brackets always means "all". It's like a function parameter, but at type level—it's in the caller's hands what it gets bound to. You couldn't write a function `func foo<Input, Output>(Input) -> Output` unless that function was able to generate a value of every possible type a caller might pass in for Output, just like you couldn't write e.g. 'absolute value' as taking its result as a second parameter.
>> 
>> >
>> > If the interpretation for output is that at time 1 it returns one of all the possible implementations at at time 2  returns another - then that is what I want. However I would describe that as returning one of the possible implementations, not all.
>> >
>> > But no doubt you are correct, since you probably wrote that bit of the compiler :(
>> >
>> > More practical points
>> >
>> >  1. My compiler, 7.3 beta (7D152p), rejects the syntax, it doesn't like where inside Any<> saying it expects > to complete generic argument list. When will this be available, so that I can try it out?
>> >  2. Will the declarations inside protocols also change to the Any<...> form or will all the generics remain following the function name rather in a returned Any<...>? Currently -> Any<...> doesn't work in protocols for me.
>> >  3. In the construct Any<Protocol where Type == Type>, i.e. same type name used in protocol and enclosing struct/class/extension, does the left Type refer to the protocol or the enclosing struct/class/extension?
>> >  4. Is there any documentation of all of this?
>> 
>> Sorry, this is all possible future syntax and features. It's not implemented today. You'd need to write your own equivalent of the "Any" wrapper by hand right now.
>> 
>> -Joe
> 
> _______________________________________________
> swift-evolution mailing list
> swift-evolution at swift.org
> https://lists.swift.org/mailman/listinfo/swift-evolution
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <https://lists.swift.org/pipermail/swift-evolution/attachments/20160310/754d42b2/attachment.html>