[swift-users] Weird protocol behaviour.

Fri Dec 23 07:03:16 CST 2016

I’m not sure what you mean. P is an existential for x: P = … where it is upgraded to the static P in foo<A : P>, but the static P does not conform to the existential P, which results in an error described by the OP.

x: P = … here it’s Any any type that conforms to P. Any is a type of it’s own, and when existentials will be implemented I’d assume that there won’t be any upgrade to P from generic context.

let x: Any<P> = …

foo(x) 
// A == Any which should make foo equivalent to 
// `func foo(_ x: Any)` or the current `func foo(_ x: P)`
Right now we just cannot be explicit about existentials.

Please correct me if I’m wrong. ;)

-- 
Adrian Zubarev
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Am 23. Dezember 2016 um 13:47:18, Zhao Xin (owenzx at gmail.com) schrieb:

You mis-labelled you problem in your original email.

    let x = X() as P // this does not compile. Why?
    foo(x)

The issue is not in line ` let x = X() as P`. It is in line ` foo(x)`, `x's type is P`, but `foo(:)` request a parameter type of `A`, not `P`.

`func foo<A:P>(_ x:A) {}` means, `x` must be `A`, and `A` conforms `P`. But not all `P`s are `A`.

Zhaoxin

On Fri, Dec 23, 2016 at 5:26 PM, Adrian Zubarev via swift-users <swift-users at swift.org> wrote:
Whoops, wait a second. Correcting my self here. Copied the wrong url.

Here is the proposal I meant. LINK

-- 
Adrian Zubarev
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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)?!

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
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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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