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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Ah, one should never post to a mailing list before a morning coffee! The
reason for the essentially undefined behaviour is that `Hashable`
conformance must guarantee that two instances have the same `hashValue`
if they are `==`, **but not vice versa**! In other words, the equality
in the code s
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 (8A218a
