[swift-evolution] two protocols with the same method name

[肇鑫]

Howard is almost right.

Protocol A means the ring variable must have a getter. For the setter, you
can have it or not. Both won't violate the conformance of the Protocol A.

Actually, you can write as

protocol B:A {
var ring:String { get set }
}

The is more for the last line
// (x as A).ring = "🔔" Error, A doesn't have set

You will find more that
(x as B).ring = "1"
doesn't work either. The reason is that (x as B) is a let by default.

You need do
var b = x as B
b.ring = "1"

Above works.

zhaoxin

On Sat, Jan 9, 2016 at 10:19 AM, Howard Lovatt via swift-evolution <
swift-evolution at swift.org> wrote:

> I don’t really get what you are driving at. If I rework your example so
> that it runs:
>
> //💍
> protocol A {
>     var ring: String { get }
> }
>
> //🔔
> protocol B {
>     var ring: String { get set }
> }
>
> class X: A, B {
>     var _ring = "💍"
>     var ring: String {
>         get {
>             return _ring
>         }
>         set {
>             self._ring = newValue
>         }
>     }
> }
> let x = X()
> (x as A).ring // 💍
> (x as B).ring // 💍
> x.ring // 💍
> x.ring = "🔔"
> x.ring // 🔔
> // (x as A).ring = "🔔" Error, A doesn't have set
>
> Then the behaviour is exactly what I would expect.
>
> On 7 Jan 2016, at 9:18 PM, Grzegorz Leszek via swift-evolution <
> swift-evolution at swift.org> wrote:
>
> I suggest compile warning if one classes/structs/enums implements
> protocols with the same name.
> It could lead to confusions, when methods of those protocols will mean
> different things.
> It will force to implement parent protocol with shared methods or
> change method in one of the protocols.
> Below is an example.
> Regards,
> Greg
>
> //💍
> protocol A {
>  var ring: String { get }
> }
>
> //🔔
> protocol B {
>  var ring: String { get set }
> }
>
> class X: A, B {
>  var ring: String {
>    get {
>      return "💍"
>    }
>    set {
>      self.ring = newValue
>    }
>  }
> }
> let x = X()
> let somewhereInTheProject = "\(x.ring) the bell"
> x.ring = "🔔" // ERROR!
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>


-- 

Owen Zhao
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