How to compare enum with associated values by ignoring its associated value in Swift?

Solution 1:

Edit: As Etan points out, you can omit the (_) wildcard match to use this more cleanly.

Unfortunately, I don't believe that there's an easier way than your switch approach in Swift 1.2.

In Swift 2, however, you can use the new if-case pattern match:

let number = CardRank.Number(5)
if case .Number(_) = number {
    // Is a number
} else {
    // Something else
}

If you're looking to avoid verbosity, you might consider adding an isNumber computed property to your enum that implements your switch statement.

Solution 2:

Unfortunately in Swift 1.x there isn't another way so you have to use switch which isn't as elegant as Swift 2's version where you can use if case:

if case .Number = number {
    //ignore the value
}
if case .Number(let x) = number {
    //without ignoring
}

Solution 3:

In Swift 4.2 Equatable will be synthesized if all your associated values conform to Equatable. All you need to do is add Equatable.

enum CardRank: Equatable {
    case Number(Int)
    case Jack
    case Queen
    case King
    case Ace
}

https://developer.apple.com/documentation/swift/equatable?changes=_3

Solution 4:

Here's a simpler approach:

enum CardRank {
    case Two
    case Three
    case Four
    case Five
    case Six
    case Seven
    case Eight
    case Nine
    case Ten
    case Jack
    case Queen
    case King
    case Ace

    var isFaceCard: Bool {
        return (self == Jack) || (self == Queen) || (self == King)
    }
}

There's no need to overload the == operator, and checking for card type does not require confusing syntax:

let card = CardRank.Jack

if card == CardRank.Jack {
    print("You played a jack")
} else if !card.isFaceCard {
    print("You must play a face card!")
}