[swift-evolution] Method dispatching issue with subclasses implementing Equatable protocol.

[Pierre Monod-Broca]

The way I understand it, it's a bad idea to override == and != (or any infix operator) for Sub if Super has them and that's why the default implementation from Equatable only generates !=(Super, Super) and not !=(Sub, Sub) (and there is no ==(Sub, Sub) generated either).

And it's a bad idea because (without dynamic dispatch on both operands) it leads to unexpected behavior.

Considering :
```
func ==(lhs: Super, rhs: Super) -> Bool {
    print("Super")
    return true
}

func ==(lhs: Sub, rhs: Sub) -> Bool {
    print("Sub")
    return false
}

let a = Sub()
let b = Sub()
a == b // Sub
a as Super == b // Super
a == b as Super // Super
à as Super == b as Super // Super
```

One would compare the same objects and don't get the same result.

Instead you have to check the dynamic type yourself.


Pierre

> Le 20 janv. 2017 à 10:45, Francisco Javier Fernández Toro via swift-evolution <swift-evolution at swift.org> a écrit :
> 
> 
> 
>> On Wed, Jan 18, 2017 at 6:58 PM, Tony Allevato <tony.allevato at gmail.com> wrote:
>> Ok, this actually does feel a bit strange. The behavior you're seeing seems to be a consequence of [SE-0091](https://github.com/apple/swift-evolution/blob/master/proposals/0091-improving-operators-in-protocols.md), but it looks like you're seeing different behavior than what I described in the "Class types and inheritance" section of that proposal.
>> 
>> If Sub has `==(Sub, Sub)` implemented as a *static* function, I just tried it and it's *ignored* (`==(Super, Super)` gets called instead), even when the two actual arguments are known to be statically of type Sub. I think this is because of the way that proposal was implemented: when it sees that `Sub` extends `Super`, which conforms to `Equatable`, it appears that it's only looking for static overloads of `==` that are satisfied at the *point of conformance*, which would be `==(Super, Super)` (because `Super` conforms to `Equatable where Self == Super`). The wording of the proposal makes this case: "Then, we say that we do not consider an operator function if it implements a protocol requirement, because the requirement is a generalization of all of the operator functions that satisfy that requirement."
>> 
>> Contrarily, if you provide `==(Sub, Sub)` as a global function instead of a static one, it *does* get called. I think in this case, the type checker gets the whole set of candidate operators (which, unlike above, includes the global `==(Sub, Sub)`), and it gets used because it's a more specific match?
>> 
> 
> FWIW, I've just changed both `==` functions to make them global, the the outcome is still the same, its using `==(Super,Super)` to resolve `!=(Sub,Sub)
>  
>> Can someone from the core team chime in and say whether this is intentional behavior? It feels wrong that simply changing the location where the operator is defined would change the behavior like this.
>> 
>> FWIW, to avoid these sharp edges, there's no need to implement `==` for subtypes; since you have to use an overridable `equals` method anyway, just have the base type implement `==` to delegate to it, and then have subtypes override `equals` alone.
>> 
>> 
>>> On Wed, Jan 18, 2017 at 9:36 AM Francisco Javier Fernández Toro <fran at gokarumi.com> wrote:
>>> Yeah guys, you are right, my code is busted, I was trying to point something different out:
>>> 
>>> The next code is showing the possible issue. In theory to make a class Equatable, you just have to mark it with the Equatable protocol and implement `==` as a static function or as a global one.
>>> 
>>> If you don't override the equal method and you just invoke your super class equality method you'll get something like this: 
>>> 
>>> ```
>>> class Superclass : Equatable {
>>>     let foo: Int
>>>     
>>>     init(foo: Int) { self.foo = foo }
>>>     
>>>     func equal(to: Superclass) -> Bool {
>>>         return foo == to.foo
>>>     }
>>>     
>>>     static func == (lhs: Superclass, rhs: Superclass) -> Bool {
>>>         return lhs.equal(to: rhs)
>>>     }
>>> }
>>> 
>>> class Subclass: Superclass {
>>>     let bar: Int
>>>     init(foo: Int, bar: Int) {
>>>         self.bar = bar
>>>         super.init(foo: foo)
>>>     }
>>>     
>>>     func equal(to: Subclass) -> Bool {
>>>         return bar == to.bar && super.equal(to: to)
>>>     }
>>>     
>>>     static func == (lhs: Subclass, rhs: Subclass) -> Bool {
>>>         return lhs.equal(to: rhs)
>>>     }
>>> }
>>> 
>>> class SubclassWithDifferentOperator: Subclass {
>>>     static func != (lhs: SubclassWithDifferentOperator, rhs: SubclassWithDifferentOperator) -> Bool {
>>>         return !(lhs.equal(to: rhs))
>>>     }
>>> }
>>> 
>>> let a = Subclass(foo: 1, bar: 1)
>>> let b = Subclass(foo: 1, bar: 2)
>>> 
>>> (a == b) != (a != b) // Prints: false, not expected
>>> 
>>> let x = SubclassWithDifferentOperator(foo: 1, bar: 1)
>>> let y = SubclassWithDifferentOperator(foo: 1, bar: 2)
>>> 
>>> (x == y) != (x != y) // Prints: true, expected
>>> ```
>>> 
>>> So, after adding a couple of `print` statement in those equal method what I can see is that for Subclass, when you are need to call `!=` what Swift is doing is using `func ==(Superclass, Superclass)` and apply `!` as Tony has pointed out.
>>> 
>>> What I cannot understand is why is not using `func == (Subclass, Subclass)`
>>> 
>>> I hope it makes more sense now.
>>> 
>>> ---
>>> Fran Fernandez
>>> 
>>> On Wed, Jan 18, 2017 at 6:13 PM, Tony Allevato <tony.allevato at gmail.com> wrote:
>>> This seems to work for me:
>>> 
>>> ```
>>> class Super: Equatable { 
>>>     let x: Int
>>>     init(x: Int) {
>>>         self.x = x
>>>     }
>>>     func equals(_ rhs: Super) -> Bool { 
>>>         return x == rhs.x 
>>>     } 
>>>     static func ==(lhs: Super, rhs: Super) -> Bool { 
>>>         return lhs.equals(rhs) 
>>>     } 
>>> } 
>>> 
>>> class Sub: Super {
>>>     let y: Int
>>>     init(x: Int, y: Int) {
>>>         self.y = y
>>>         super.init(x: x)
>>>     }
>>>     override func equals(_ rhs: Super) -> Bool {
>>>         if let rhs = rhs as? Sub {
>>>             return y == rhs.y && super.equals(rhs)
>>>         }
>>>         return false
>>>     }   
>>> }
>>> 
>>> let a = Sub(x: 1, y: 1)
>>> let b = Sub(x: 1, y: 2)
>>> let c = Sub(x: 1, y: 1)
>>> 
>>> a == b  // false, expected
>>> a == c  // true, expected
>>> a != b  // true, expected
>>> a != c  // false, expected
>>> ```
>>> 
>>> Additionally, when I made the change Joe suggested, your code also worked, so maybe there was an error when you updated it?
>>> 
>>> FWIW, the default implementation of != just invokes !(a == b) <https://github.com/apple/swift/blob/master/stdlib/public/core/Equatable.swift#L179-L181>, so I believe it's *impossible* (well, uh, barring busted RAM or processor I guess) for it to return the wrong value for the same arguments if you only implement ==.
>>> 
>>> 
>>> 
>>> On Wed, Jan 18, 2017 at 8:52 AM Francisco Javier Fernández Toro via swift-evolution <swift-evolution at swift.org> wrote:
>>> Thank you for your answer Joe,
>>> 
>>> you are right the equal(to:) wasn't a valid override, but even after using the one you've proposed, the behavior is not the expected one
>>> 
>>> 
>>> let a = Subclass(foo: 1, bar: 1)
>>> let b = Subclass(foo: 1, bar: 2)
>>> 
>>> (a == b) != (a != b) // Prints true
>>> 
>>> let x = SubclassWithDifferentOperator(foo: 1, bar: 1)
>>> let y = SubclassWithDifferentOperator(foo: 1, bar: 2)
>>> 
>>> (x == y) != (x != y) // Prints false
>>> 
>>> As you can see above if a subclass does not implement the global function !=, the equal operation seems to be broken.
>>> 
>>> ---
>>> 
>>> Fran Fernandez
>>> 
>>> On Wed, Jan 18, 2017 at 5:44 PM, Joe Groff <jgroff at apple.com> wrote:
>>> 
>>> > On Jan 18, 2017, at 2:59 AM, Francisco Javier Fernández Toro via swift-evolution <swift-evolution at swift.org> wrote:
>>> >
>>> > Hi,
>>> >
>>> > I've found that when you have a class hierarchy which implements Equatable, if you want to have the != operator working as expected, you need to override it, it's not enough with ==.
>>> >
>>> > If you don't define you own subclass != operator, Swift compiler will use the super class to resolve that operation.
>>> >
>>> > Is there any reason for that?
>>> 
>>> The `equal(to:)` method inside `Subclass` is not a valid override of `Superclass` because its argument only accepts `Subclass` instances, but the parent method needs to work with all `Superclass` instances. If you write it as an override, it should work:
>>> 
>>> class Subclass: Superclass {
>>>     let bar: Int
>>>     init(foo: Int, bar: Int) {
>>>         self.bar = bar
>>>         super.init(foo: foo)
>>>     }
>>> 
>>>     override func equal(to: Superclass) -> Bool {
>>>       if let toSub = to as? Subclass {
>>>         return bar == toSub.bar && super.equal(to: to)
>>>       }
>>>       return false
>>>     }
>>> }
>>> 
>>> We should probably raise an error, or at least a warning, instead of silently accepting your code as an overload. Would you be able to file a bug on bugs.swift.org about that?
>>> 
>>> -Joe
>>> 
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>>> swift-evolution at swift.org
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>>> 
> 
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