Hello Jeremy,
It seems it is a little bit more complex than this:
http://allblue.me/swift/2016/01/23/swift-method-dispatch-with-protocol-extension-protocol-extension-and-subclass/
[image: diagram]
- IF the inferred type of a variable is the protocol:
- IF the method is in a class conform directly to the original
protocol:
- AND the method is defined in the original protocol
- THEN the runtime type’s implementation is called,
irrespective of whether there is a default implementation
in the extension.
- AND the method is not defined in the original protocol,
- THEN the default implementation is called.
- ELSE IF the method is in a subclass of another class who conform
to the original protocol
- THEN the default implementation of protocol is called.
- ELSE IF the inferred type of the variable is the type
- THEN the type’s implementation is called.
On Thu, May 19, 2016 at 5:24 PM, Jeremy Pereira via swift-evolution <
[email protected]> wrote:
>
> > On 19 May 2016, at 13:30, Krystof Vasa via swift-evolution <
> [email protected]> wrote:
> >
> > See this example that demonstrates how it's pretty much unusable (IMHO),
> since whenever you refer to the instance as to the protocol, the default
> implementation gets invoked:
> >
> > protocol MyProtocol { }
> >
> > extension MyProtocol {
> > func getInt() -> Int {
> > return 0
> > }
> > }
> >
> > class MyClass: MyProtocol {
> > func getInt() -> Int {
> > return 1
> > }
> > }
> >
> >
> > let instance = MyClass()
> > instance.getInt() // 1
> >
> > var anyInstance: MyProtocol = instance
> > anyInstance.getInt() // 0 !!!!!!!
> >
> >
> > Since anyInstance is of MyProtocol type, you get the default
> implementation (no dynamic dispatch).
>
>
> That’s because the only information that the compiler has about
> anyInstance is that it conforms to MyProtocol which has no methods so it
> doesn’t know that it can dispatch getInt() to the implementation in
> MyClass. Change the protocol to
>
> protocol MyProtocol {
> func getInt() -> Int
> }
>
> and it will work as expected.
>
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>
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