The == operator is defined for tuples with (up to 6) elements that conform to
the Equatable protocol
(and < for tuples with Comparable elements):
class SomeClass: Equatable {
static public func ==(_ lhs:SomeClass, _ rhs:SomeClass) -> Bool {
return lhs === rhs
}
}
let c1 = SomeClass()
let c2 = SomeClass()
let t1 = ("abc", c1)
let t2 = ("abc", c2)
c1 == c2 // legal
t1 == t2 // legal
The existence of a == operator alone, without declaring protocol conformance,
is not sufficient.
> On 9. Jul 2017, at 17:11, David Baraff via swift-users
> <swift-users@swift.org> wrote:
>
> Given 2-tuples of type (T1, T2), you should be able to invoke the == operator
> if you could on both types T1 and T2, right? i.e.
>
> (“abc”, 3) == (“abc”, 4) // legal
>
> but:
>
> class SomeClass {
> static public func ==(_ lhs:SomeClass, _ rhs:SomeClass) -> Bool {
> return lhs === rhs
> }
> }
>
> let c1 = SomeClass()
> let c2 = SomeClass()
>
> let t1 = ("abc", c1)
> let t2 = ("abc", c2)
>
> c1 == c2 // legal
> t1 == t2 // illegal
>
>
>
>
> Why is t1 == t2 not legal given that c1 == c2 IS legal?
>
>
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> swift-users@swift.org
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