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<P> any type that conforms to P. Any<P> 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.
Please correct me if I’m wrong. ;)
···
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Adrian Zubarev
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Am 23. Dezember 2016 um 13:47:18, Zhao Xin (owenzx@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@swift.org> wrote:
Whoops, wait a second. Correcting my self here. Copied the wrong url.
Here is the proposal I meant. LINK
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Adrian Zubarev
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Am 23. Dezember 2016 um 09:57:49, Adrian Zubarev (adrian.zubarev@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.
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Adrian Zubarev
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Am 23. Dezember 2016 um 09:43:34, Mikhail Seriukov via swift-users (swift-users@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@starbase55.com>:
On 22 Dec 2016, at 22:43, Howard Lovatt <howard.lovatt@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@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@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@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: Swiftrien (Rien) · GitHub
Project: http://swiftfire.nl
On 22 Dec 2016, at 17:05, Mikhail Seriukov via swift-users <swift-users@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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