[swift-users] Weird protocol behaviour.

[Mikhail Seriukov]

>
> 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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>>
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>>
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>>
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>>
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>>
>
>
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