Btw,..

On 20.09.2016 18:18, Nevin Brackett-Rozinsky via swift-evolution 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.

See how "my" suggestion regarding requiring a keyword for implementation of protocol requirement (but with `override` keyword instead of `implement`) can be very helpful here (And Karl, seems like "your" suggestion will not help here - I mean we'll have to change bar() definition in B in case A implement its own bar())

In this case, initially, B.bar() must be marked with "override" as it is implementing protocol's requirement. *And* this will "protect" B class in future if class A will implement its own bar() method:

protocol Foo {
        func bar()
}

extension Foo {
        func bar() {
                print("I am bar.")
        }
}

class A:Foo {
}

class B:A {
        *override* func bar() { // currently protocol impelentation
                print("I am B.")
        }
}

and then if

class A:Foo {
        override func bar() {...}
}

,B's definition will not be changed.

Then we have a question regarding super.bar() inside B.bar(), i.e. should it be accessible or not. I believe we should consider if it could be implemented, so in case A has no bar(), super.bar() will call protocol's implementation, otherwise (if A contains bar()) - A.bar() will be called. Don't know if this is what very hard to implement. Probably as first step super.bar() will not be allowed inside of protocol implementation.

Swift allows for protocols to have default *implementations* for requirements. And *also* Swift allows you to *not* implement requirement that has default implementation in conformed type. When you don't implement such requirement in type - the type *inherits* that implementation. We have classical inheritance scheme like in inheritance of super type's method. When you override super type's method you have to use `override`. I believe overriding the inherited default implementation of requirement - also should be marked with `override`.

Then, Swift allows protocol to have extension in any source file in project. You can have no default implementation *at moment of writing* your type, but you can have such default implementation *at moment of compilation*. So, as soon as we *can't* know if protocol's requirement will *actually* have default implementation during the compilation - logically that we should use `override` to mark each method/prop defined as implementation of protocol, to suppose that there *is* default implementation which will be *overridden* by our custom implementation.


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
<swift-evolution@swift.org <mailto:swift-evolution@swift.org>> 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
    <swift-evolution@swift.org <mailto:swift-evolution@swift.org>> 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
        (swift-us...@swift.org <mailto:swift-us...@swift.org>) 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

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