[swift-users] Attempting to call default protocol implementation crashes Playground

[Rick Mann]

Well, this is a standard protocol default implementation. I was experimenting to see if it was possible to call the default implementation after providing a concrete implementation.

> On Nov 15, 2016, at 14:47 , Dan Loewenherz <dan at lionheartsw.com> wrote:
> 
> What are you trying to accomplish here, more concretely?
> 
> My first thought is that you shouldn't implement the same function in both a protocol extension and a conforming class. Why not just give them different names and call the function from within the extension instead of from the class? E.g.
> 
> protocol FooPro {
>     func _fooFunc()
> }
> 
> extension FooPro {
>     func fooFunc() {
>         print("fooFunc default")
>         _fooFunc()
>     }
> }
> 
> class FooClass: FooPro {
>     func _fooFunc() {
>         print("fooFunc FooClass")
>     }
> }
> 
> let fc = FooClass()
> fc.fooFunc()
> 
> Dan
> 
> On Tue, Nov 15, 2016 at 4:28 PM, Rick Mann via swift-users <swift-users at swift.org> wrote:
> The following gives Xcode 8.1 a very hard time. Eventually I get a Bad Access on the last line. I'm guessing it's a recursive call. Is there any way to call the default implementation from a "real" implementation?
> 
> protocol FooPro
> {
>         func fooFunc()
> }
> 
> extension FooPro
> {
>         func
>         fooFunc()
>         {
>                 print("fooFunc default")
>         }
> }
> 
> class FooClass : FooPro
> {
>         func
>         fooFunc()
>         {
>                 (self as FooPro).fooFunc()
>                 print("fooFunc FooClass")
>         }
> }
> 
> let fc: FooPro = FooClass()
> fc.fooFunc()
> 
> 
> Thanks!
> 
> 
> --
> Rick Mann
> rmann at latencyzero.com
> 
> 
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> 


-- 
Rick Mann
rmann at latencyzero.com