Yes, Rien, just realised that. Now I must find some place to hide in
shame…
Thanks for your reply, though!
milos
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If you change the “==“ function to:
func == (lhs: X, rhs: X) -> Bool {
return (lhs.x == rhs.x) && (lhs.hashValue == rhs.hashValue)
}
then the example works as expected.
Apple says this: “A hash value, provided by a type’s hashValue property, is an
integer that is the same for any two
Given an array of instances of a `Hashable` value type, all equal
according to `Equatable` protocol, but with distinct `hashValue`s, I
would expect that initialising a set with that array would preserve all
the instances. Instead, running the code below in an iOS playground on
Xcode 8.0
