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 <jgr...@apple.com> wrote:
>
> > On Jan 18, 2017, at 2:59 AM, Francisco Javier Fernández Toro via
> swift-evolution <swift-evolution@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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