[swift-users] Weird protocol behaviour.

[Adrian Zubarev]

I assume when we get existentials, your problem could be solved like this:

protocol P {}
class X: P {}

func foo<A:P>(_ x:A) {}

func bar() {
    let x = X() as Any<P>
    foo(x)
}
Here A will be Any<P> which conforms to P and makes the compiler happy.

let c: P = … here is P and existential like (future) Any<P>.
x as P here is P used as a type.
It’s weird that the latter is a type but the first example is an existential. I think this is a design choice, because we almost never need the protocol as a type.



-- 
Adrian Zubarev
Sent with Airmail

Am 23. Dezember 2016 um 09:57:49, Adrian Zubarev (adrian.zubarev at devandartist.com) schrieb:

What are you trying to solve here?

Do you heavily rely on what A can be?

Can’t you just write func foo(_ x: P) {} (here P is an existential like [in some future] Any<P>)?!

AnyObject is for classes, but you’d get the same result changing X to a class ;)

If you want scratch the surface of what happens to protocols in a generic context, you could read our proposal about meta types here.



-- 
Adrian Zubarev
Sent with Airmail

Am 23. Dezember 2016 um 09:43:34, Mikhail Seriukov via swift-users (swift-users at swift.org) schrieb:

No it does not.
You have made a type out of the parameter. It’s no longer a protocol.
IMO the failure here is to understand the difference between a type and a protocol.
A type (even if empty) is always a combination of storage with functions (that are assumed to work on the data in storage)
A protocol is just a definition of functions without the accompanying data.

I see your point. 
But actually when I write it as  `let x = X() as P` I really mean that I want `x` to be `AnyObject` but conforming to P, not just protocol itself.
Is it even possible to downcast it this way?

2016-12-23 14:51 GMT+07:00 Marinus van der Lugt <rien at starbase55.com>:

On 22 Dec 2016, at 22:43, Howard Lovatt <howard.lovatt at gmail.com> wrote:

The following variation works:

protocol P {}

class P1:P {}

class X:P1 {}

func foo<A:P>(_ x:A) {}

func bar() {
    //let x = X() // this compiles
    let x = X() as P1 // this does not compile. Why?
    foo(x)
}

Which adds credence to the bug theory.


No it does not.
You have made a type out of the parameter. It’s no longer a protocol.
IMO the failure here is to understand the difference between a type and a protocol.
A type (even if empty) is always a combination of storage with functions (that are assumed to work on the data in storage)
A protocol is just a definition of functions without the accompanying data.

Rien.




Note two changes: 1. two levels of inheritance and 2. change to classes. If you do two levels using protocols it doesn't work if you use either classes or structs.


  -- Howard.

On 23 December 2016 at 07:29, Kevin Nattinger <swift at nattinger.net> wrote:
I recall seeing a request on the -evolution list for something like `T := X` to indicate it could be X itself or anything inheriting / implementing it, so it’s certainly known behavior, if not desired. IMO it’s a bug and `:` should be fixed to include the root type, whether or not that requires a discussion on -evolution.

On Dec 22, 2016, at 2:17 PM, Howard Lovatt via swift-users <swift-users at swift.org> wrote:

I suspect a compiler bug since A is a P. The equivalent in Java works:

interface P {}
class X implements P {}
 
<A extends P> void foo(A x) {}
 
void bar() {
    final P x = new X();
    foo(x);
}

-- Howard. 

On 23 Dec 2016, at 3:19 am, Rien via swift-users <swift-users at swift.org> wrote:

IMO the error message says it all:

Playground execution failed: error: MyPlayground8.playground:9:5: error: cannot invoke 'foo' with an argument list of type '(P)'
   foo(x)
   ^

MyPlayground8.playground:9:5: note: expected an argument list of type '(A)'
   foo(x)
   ^

I.e. you are passing in a protocol while the function is specified for a type.
Said other way: On which data do you expect the protocol to operate?

Regards,
Rien

Site: http://balancingrock.nl
Blog: http://swiftrien.blogspot.com
Github: http://github.com/Swiftrien
Project: http://swiftfire.nl




On 22 Dec 2016, at 17:05, Mikhail Seriukov via swift-users <swift-users at swift.org> wrote:

Hello community! I' wondering if somebody can explain this to me.
Please take look at the snippet.

protocol P {}
struct X:P {}

func foo<A:P>(_ x:A) {}

func bar() {
   //let x = X() // this compiles
   let x = X() as P // this does not compile. Why?
   foo(x)
}

I expect the both cases to work though. But only first works? And I do not understand why.
My coworkers said that it is a compiler bug, but I'm not shure it is.
Thanks for the help.
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