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

[Pierre Monod-Broca via swift-evolution]

It's worth mentioning that with equal(to:) the dispatch is dynamic only on 
`self` not on `to`, which means different implementation can be called if you 
swap the operands. For exemple:
```
let a = Super()
let b = Sub()

a.equal(to: b) // calls Super.equal(to:)
b.equal(to: a) // calls Sub.equal(to:)
```

Pierre

> Le 21 janv. 2017 à 08:20, Karl Wagner  a écrit :
> 
> This is very interesting and all Swift developers will need to be aware of 
> it, if it isn’t a bug.
> 
> One thing that I think is a little worrying is that if, rather than the == 
> operator, you had a CustomEquatable protocol which defined the equal(to:) 
> function, that function would be dynamically dispatched. You would think that 
> operator witnesses should be resolved in the same way.
> 
> I believe that should even work for the dynamic Self; we have already 
> verified at compile-time that an overload exists (even though it may not be 
> the most appropriate one to execute). So you would get:
> 
> let a = Sub()
> let b = Sub()
> let c = Super()
> 
> (a as Super) == b // Both values have (dynamic) type Sub, calls ==(Sub,Sub)
> a == c   // One type is not a Sub, must compare using 
> ==(Super, Super)
> 
> - Krl
> 
>> On 20 Jan 2017, at 20:24, Pierre Monod-Broca via swift-evolution 
>>  wrote:
>> 
>> 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  a écrit :
>>> 
>>> 
>>> 
 On Wed, Jan 18, 2017 at 6:58 PM, Tony Allevato  
 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 
>  wrote:
> Yeah guys, you are right, my code is busted, I was trying to point 
> 

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

[Karl Wagner via swift-evolution]

This is very interesting and all Swift developers will need to be aware of it, 
if it isn’t a bug.

One thing that I think is a little worrying is that if, rather than the == 
operator, you had a CustomEquatable protocol which defined the equal(to:) 
function, that function would be dynamically dispatched. You would think that 
operator witnesses should be resolved in the same way.

I believe that should even work for the dynamic Self; we have already verified 
at compile-time that an overload exists (even though it may not be the most 
appropriate one to execute). So you would get:

let a = Sub()
let b = Sub()
let c = Super()

(a as Super) == b // Both values have (dynamic) type Sub, calls ==(Sub,Sub)
a == c   // One type is not a Sub, must compare using ==(Super, 
Super)

- Krl

> On 20 Jan 2017, at 20:24, Pierre Monod-Broca via swift-evolution 
>  wrote:
> 
> 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 
> > a écrit :
> 
>> 
>> 
>> On Wed, Jan 18, 2017 at 6:58 PM, Tony Allevato > > 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 
>> > 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: 

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

[Goffredo Marocchi via swift-evolution]

Ah... we will be missing hiding behind the warm embrace of message 
passing/dynamic dispatch :P

Sent from my iPhone

> On 20 Jan 2017, at 19:24, Pierre Monod-Broca via swift-evolution 
>  wrote:
> 
> 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  a écrit :
>> 
>> 
>> 
>>> On Wed, Jan 18, 2017 at 6:58 PM, Tony Allevato  
>>> 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 
  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 

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

[Pierre Monod-Broca via swift-evolution]

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 
>  a écrit :
> 
> 
> 
>> On Wed, Jan 18, 2017 at 6:58 PM, Tony Allevato  
>> 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 
>>>  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 

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

[Francisco Javier Fernández Toro via swift-evolution]

On Wed, Jan 18, 2017 at 6:58 PM, Tony Allevato 
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 <
> f...@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 
>> 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 {

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

[Tony Allevato via swift-evolution]

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?

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 <
f...@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 
> 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 

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

[Francisco Javier Fernández Toro via swift-evolution]

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 
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  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  wrote:
>>
>>
>> > On Jan 18, 2017, at 2:59 AM, Francisco Javier Fernández Toro via
>> swift-evolution  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)
>>  

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

[Tony Allevato via swift-evolution]

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  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  wrote:
>
>
> > On Jan 18, 2017, at 2:59 AM, Francisco Javier Fernández Toro via
> swift-evolution  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
>
>
> ___
> swift-evolution mailing list
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>
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Re: [swift-evolution] Method dispatching issue with subclasses implementing Equatable protocol.

[Francisco Javier Fernández Toro via swift-evolution]

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  wrote:

>
> > On Jan 18, 2017, at 2:59 AM, Francisco Javier Fernández Toro via
> swift-evolution  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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Re: [swift-evolution] Method dispatching issue with subclasses implementing Equatable protocol.

[Joe Groff via swift-evolution]

> On Jan 18, 2017, at 2:59 AM, Francisco Javier Fernández Toro via 
> swift-evolution  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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