Hashing problems using a wrapper class around NSUUID as the key - swift

Question

** REWRITE **
OK, it turns out I'm really asking a different question. I understand about hashValue and ==, so that's not relevant.
I would like my wrapper class BUUID to "do the right thing" and act just like NSUUID's act in a Dictionary.
See below, where they don't.
import Foundation
class BUUID: NSObject {
init?(str: String) {
if let uuid = NSUUID(UUIDString: str) {
_realUUID = uuid
}
else {
return nil
}
}
override init() {
_realUUID = NSUUID()
}
private var _realUUID: NSUUID
override var description: String { get { return _realUUID.UUIDString } }
override var hashValue: Int { get { return _realUUID.hashValue } }
var UUIDString: String { get { print("WARNING Use description or .str instead"); return _realUUID.UUIDString } }
var str: String { get { return _realUUID.UUIDString } }
}
func ==(lhs: BUUID, rhs: BUUID) -> Bool { return lhs._realUUID == rhs._realUUID }
let a = BUUID()
let b = BUUID(str: a.str)!
print("a: \(a)\nb: \(b)")
print("a === b: \(a === b)")
print("a == b: \(a == b)")
var d = [a: "Hi"]
print("\(d[a]) \(d[b])")
let nA = NSUUID()
let nB = NSUUID(UUIDString: nA.UUIDString)!
print("na: \(nA)\nnB: \(nB)")
print("nA === nB: \(nA === nB)")
print("nA == nB: \(nA == nB)")
var nD = [nA: "Hi"]
print("\(nD[nA]) \(nD[nB])")
Results. Note that I can look up using NSUUID nB and get back what I put under nA. Not so with my BUUID.
a: 9DE6FE91-D4B5-4A6B-B912-5AAF34DB41C8
b: 9DE6FE91-D4B5-4A6B-B912-5AAF34DB41C8
a === b: false
a == b: true
Optional("Hi") nil
nA: <__NSConcreteUUID 0x7fa193c39500> BB9F9851-93CF-4263-B98A-5015810E4286
nB: <__NSConcreteUUID 0x7fa193c37dd0> BB9F9851-93CF-4263-B98A-5015810E4286
nA === nB: false
nA == nB: true
Optional("Hi") Optional("Hi")

Answer 1

Inheriting from NSObject also assumes isEqual(object: AnyObject?) -> Bool method overloading:
import Foundation
class BUUID: NSObject {
init?(str: String) {
if let uuid = NSUUID(UUIDString: str) {
_realUUID = uuid
}
else {
return nil
}
}
override init() {
_realUUID = NSUUID()
}
private var _realUUID: NSUUID
override func isEqual(object: AnyObject?) -> Bool {
guard let buuid = object as? BUUID else {
return false
}
return buuid._realUUID == _realUUID
}
override var description: String { get { return _realUUID.UUIDString } }
override var hashValue: Int { get { return _realUUID.hashValue } }
var UUIDString: String { get { print("WARNING Use description or .str instead"); return _realUUID.UUIDString } }
var str: String { get { return _realUUID.UUIDString } }
}
func ==(lhs: BUUID, rhs: BUUID) -> Bool { return lhs._realUUID == rhs._realUUID }
let a = BUUID()
let b = BUUID(str: a.str)!
print("a: \(a)\nb: \(b)")
print("a === b: \(a === b)")
print("a == b: \(a == b)")
var d = [a: "Hi"]
print("\(d[a]) \(d[b])")
let nA = NSUUID()
let nB = NSUUID(UUIDString: nA.UUIDString)!
print("na: \(nA)\nnB: \(nB)")
print("nA === nB: \(nA === nB)")
print("nA == nB: \(nA == nB)")
var nD = [nA: "Hi"]
print("\(nD[nA]) \(nD[nB])")

Answer 2

So the answer is to not make BUUID inherit from NSObject, which undercuts the Swiftiness of overriding ==.
So:
extension BUUID: Hashable {}
class BUUID: CustomStringConvertible {
// take away all 'override' keywords, nothing to override
// otherwise same as above
}
Interesting!

Answer 3

This answer is relevant to initially asked question: Why that's possible to get two key-value pairs with identical key's hashes in a dictionary
This example illustrates that keys in Dictionary can have identical hashes, but equality operation should return false for different keys:
func ==(lhs: FooKey, rhs: FooKey) -> Bool {
return unsafeAddressOf(lhs) == unsafeAddressOf(rhs)
}
class FooKey: Hashable, Equatable {
var hashValue: Int {
get {
return 123
}
}
}
var d = Dictionary<FooKey, String>()
let key1 = FooKey()
let key2 = FooKey()
d[key1] = "value1"
d[key2] = "value2"
Output
[FooKey: "value1", FooKey: "value2"]
That's definitely not good to have all keys with the same hash. In this case we are getting that worst case when search element complexity fells down to O(n) (exhaustive search). But it will work.