Swift, how to implement Hashable protocol based on object reference?

Question

I\'ve started to learn swift after Java. In Java I can use any object as a key for HashSet, cause it has default hashCode and equals based on objec

Answer 1

Swift 5 barebones

For a barebones implementation of making a class hashable, we simply must conform to both the Equatable and Hashable protocols. And if we know our object well, we can determine which property or properties to use to make it equatable and hashable.

class CustomClass: Equatable, Hashable {
    let userId: String
    let name: String
    let count: Int
    
    init(userId: String, name: String, count: Int) {
        self.userId = userId
        self.name = name
        self.count = count
    }
    
    /* Equatable protocol simply means establishing a predicate to
       determine if two instances of the same type are equal or unequal based
       on what we consider equal and unequal. In this class, userId makes
       the most sense so if we were to compare two instances (left-hand-side
       versus right-hand-side), we would compare their userId values. When
       it comes time to compare these objects with each other, the machine
       will look for this function and use it to make that determination. */
    static func == (lhs: CustomClass, rhs: CustomClass) -> Bool {
        return lhs.userId == rhs.userId
    }
    
    /* Hashable protocol is similar to Equatable in that it requires us to
       establish a predicate to determine if two instances of the same
       type are equal or unequal, again based on what we consider equal and
       unequal. But in this protocol we must feed that property into a
       function which will produce the object's hash value. And again, userId
       makes the most sense because each instance carries a unique value. If
       userId was not unique, we could combine multiple properties to
       generate a unique hash. */
    func hash(into hasher: inout Hasher) {
        hasher.combine(userId)
        //hasher.combine(name) if userId was not unique, we could have added this
    }
}