I understand your point on "adding additional `override` to re-implementation of default method in protocol extension". However, I don't stand for it.
If you use `override`, that means `B` could use `super.bar()` to get the original `bar()` implementation. Then if `class A` implement `bar()` someday. The `super.bar()` call in `class B` will be ambitious, it could be `A.bar()` or `Foo.bar()`. That ambiguity is the same as the multiple object inherences languages like c++. We all know that Swift can only inherit to one object. That is why the behavior is what it is now. Zhaoxin On Tue, Sep 20, 2016 at 11:18 PM, Nevin Brackett-Rozinsky < [email protected]> wrote: > I think there is a deeper issue that may be worth exploring here. > > Notably, when one class presents a member function, its subclasses ought > to use “override” when they reimplement that method themselves, regardless > of where the superclass’s version comes from. > > In the original post, the class “A” expresses (by conforming to protocol > “Foo”) that it has a member function “bar()”, and “B” is a subclass of “A” > which wants its own definition of “bar()”. > > It seems to me that “B” should not care whether “A” rolled its own > implementation of “bar()” or used the default implementation provided by > “Foo”. > > From the perspective of “B”, its superclass “A” promises to have a member > function “bar()”, so “B” should need to use the `override` keyword just > like it would when overriding any other method. > > To illustrate this more clearly, suppose that “Foo” and “A: Foo” are > defined in a 3rd-party library, while “B: A” is written in a client module. > > If the library changes to give “A” its own custom implementation of > “bar()”, that *should not* affect client code—because the class “A” still > conforms to “Foo” so it still is known to have a “bar()” method—but right > now it *does*: > > With the status quo, the simple change of moving a function between a > protocol extension and a conforming class currently requires downstream > source-code modifications in clients (in this case, adding `override` to > “B.bar()”). > > I propose that `override` should be required in subclasses on any method > which the superclass proffers as a customization point, no matter the > provenance of that claim. > > Nevin > > > > On Tue, Sep 20, 2016 at 8:01 AM, Zhao Xin via swift-evolution < > [email protected]> wrote: > >> Thank you to Игорь Никитин and Adrian Zubarev. Now I can convince myself >> for this behavior. >> >> As Adrian's example shows, there is no `bar()` implemented in `class A`, >> so there is no `override func bar()` in class B`. My thought that >> `self.bar()` >> would do the same as `(self as! B).bar()`, is called "dynamic binding", >> which is basing on the `override`. Since there is no `override`, there is >> no "dynamic binding". I thought "dynamic binding" was basing on dynamic >> type of `self`. It was not. I was wrong. >> >> The behavior is clear now. In class A's `self.bar()`, the runtime finds >> that there is no implementation of `bar()` in `class A`, so it calls the >> `bar` in protocol extension. In class A's `(self as! B).bar()`, as `class >> B` contains the implementation of `bar()`, the runtime calls it. >> >> Zhaoxin >> >> On Tue, Sep 20, 2016 at 4:21 PM, Adrian Zubarev via swift-evolution < >> [email protected]> wrote: >> >>> I can’t tell you the reason, but to me it feels like it’s doing the >>> following thing: >>> >>> + - - (Type: B) - - + >>> | | >>> | func bar() + - + (Type: A) - - + < - - - - - - >>> - - - - - - - - - - - - - - + >>> | | | >>> | >>> + - - - - - - - + func output() + - + (Protocol: Foo) - - + — - self.bar() >>> - + - (self as! B).bar() - + >>> | | | >>> | >>> + - - - - - - - + (default) func bar() | <- - - - - - - >>> - + >>> | | >>> + - - - - - - - - - - - - + >>> >>> class A:Foo { >>> >>> func bar() {} >>> >>> func output() { >>> print(type(of:self)) >>> self.bar() >>> (self as! B).bar() >>> } >>> } >>> >>> class B:A { >>> >>> override func bar() { >>> print("I am B.") >>> } >>> } >>> >>> Would solve this temporarily. >>> >>> And there we are again with the same discussion if custom implementation >>> of protocol members, which have default implementation, should have the >>> override keyword or not. >>> >>> Imagine your code like this (not valid code): >>> >>> protocol Foo { >>> func bar() >>> } >>> >>> extension Foo { >>> func bar() { >>> print("I am bar.") >>> } >>> } >>> >>> class A : Foo { >>> >>> func output() { >>> >>> print(type(of:self)) >>> default.bar() // fallback an call directly the default >>> implementation whenever needed >>> self.bar() // will print "I am bar." on A().output() but should >>> print "I am B." if Self == B >>> (self as! B).bar() >>> } >>> } >>> >>> class B : A { >>> >>> override func bar() { >>> print("I am B.") >>> } >>> } >>> >>> I still think default implementations should be called through something >>> like default. + whenever you override a default implementation you’d >>> need override. There is a discussion going on: Mark protocol methods >>> with their protocol. I clearly did not solved your issue, but I might >>> have wake your interest to participate. ;) >>> >>> >>> >>> -- >>> Adrian Zubarev >>> Sent with Airmail >>> >>> Am 20. September 2016 um 04:13:22, Zhao Xin via swift-users ( >>> [email protected]) schrieb: >>> >>> See below code. >>> >>> protocol Foo { >>> >>> func bar() >>> >>> } >>> >>> >>> extension Foo { >>> >>> func bar() { >>> >>> print("I am bar.") >>> >>> } >>> >>> } >>> >>> >>> class A:Foo { >>> >>> func output() { >>> >>> print(type(of:self)) // prints "B". >>> >>> self.bar() // prints "I am bar." >>> >>> (self as! B).bar() // prints "I am B." >>> >>> } >>> >>> } >>> >>> >>> class B:A { >>> >>> func bar() { >>> >>> print("I am B.") >>> >>> } >>> >>> } >>> >>> >>> let b = B() >>> >>> b.output() >>> >>> >>> I thought `self.bar()` would do the same as `(self as! B).bar()`. It >>> didn't. In my opinion, `type(of:self) is B.type`, so they should be the >>> same, shouldn't they? >>> >>> >>> Zhaoxin >>> _______________________________________________ >>> swift-users mailing list >>> [email protected] >>> https://lists.swift.org/mailman/listinfo/swift-users >>> >>> >>> _______________________________________________ >>> swift-evolution mailing list >>> [email protected] >>> https://lists.swift.org/mailman/listinfo/swift-evolution >>> >>> >> >> _______________________________________________ >> swift-evolution mailing list >> [email protected] >> https://lists.swift.org/mailman/listinfo/swift-evolution >> >> >
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