[swift-evolution] [swift-users] [swift-user]Unexpected behavior of protocol extension.
Goffredo Marocchi
panajev at gmail.com
Tue Sep 20 15:10:27 CDT 2016
This would make for hilarious debugging session if you have the misfortune of using a third party binary framework that gets updated and provide default implementations for some methods in the protocol extensions.
When Java 8 also decided to blur the lines between classes (implementations) and interfaces (equivalent of protocols on this side of the fence more or less... anyways they represent behaviours/API contracts), they also provided a very simple rule to determine on the code would behave at runtime.
> Extending Interfaces That Contain Default Methods
>
> When you extend an interface that contains a default method, you can do the following:
>
> Not mention the default method at all, which lets your extended interface inherit the default method.
> Redeclare the default method, which makes it abstract.
> Redefine the default method, which overrides it.
https://docs.oracle.com/javase/tutorial/java/IandI/defaultmethods.html
I am not sure why protocol extension need to differ so much and present dispatch rules that are potentially very confusing.
Casting at runtime should *never* change what implementation of a method gets called.
For a safe by default language the fact that this can occur is quite puzzling and worrying. I am not sure the cost of implementing the current static/dynamic dispatch rules for default methods in protocol extensions is worth it to say it bluntly, but please enlighten me if I am missing an obvious huge pink elephant in the room here.
Sent from my iPhone
> On 20 Sep 2016, at 16:18, Nevin Brackett-Rozinsky via swift-evolution <swift-evolution at swift.org> 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 <swift-evolution at 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 at 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-users at 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
>>>> _______________________________________________
>>>> swift-users mailing list
>>>> swift-users at swift.org
>>>> https://lists.swift.org/mailman/listinfo/swift-users
>>>
>>>
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>>
>>
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