Swift 5解决方案:

enum Suit: String, CaseIterable {
    case spades = "♠"
    case hearts = "♥"
    case diamonds = "♦"
    case clubs = "♣"
}

// access cases like this:

for suitKey in Suit.allCases {
    print(suitKey)
}

Swift 5解决方案:

enum Suit: String, CaseIterable {
    case spades = "♠"
    case hearts = "♥"
    case diamonds = "♦"
    case clubs = "♣"
}

// access cases like this:

for suitKey in Suit.allCases {
    print(suitKey)
}

在Swift中，枚举类型可以像EnumType一样访问。案例:

let tableView = UITableView(frame: self.view. view)UITableViewStyle.Plain)

大多数情况下，只有当您有几个选项可以使用，并且确切地知道在每个选项上要做什么时，才会使用枚举类型。

在处理枚举类型时，使用for-in结构没有太大意义。

你可以这样做，例如:

func sumNumbers(numbers : Int...) -> Int {
    var sum = 0

    for number in numbers{
        sum += number
    }

    return sum
}

我的解决方案是声明一个包含所有枚举可能性的数组。所以for循环可以遍历所有这些。

//Function inside struct Card
static func generateFullDeck() -> [Card] {
    let allRanks = [Rank.Ace, Rank.Two, Rank.Three, Rank.Four, Rank.Five, Rank.Six, Rank.Seven, Rank.Eight, Rank.Nine, Rank.Ten, Rank.Jack, Rank.Queen, Rank.King]
    let allSuits = [Suit.Hearts, Suit.Diamonds, Suit.Clubs, Suit.Spades]
    var myFullDeck: [Card] = []

    for myRank in allRanks {
        for mySuit in allSuits {
            myFullDeck.append(Card(rank: myRank, suit: mySuit))
        }
    }
    return myFullDeck
}

//actual use:
let aFullDeck = Card.generateFullDeck()    //Generate the desired full deck

var allDesc: [String] = []
for aCard in aFullDeck {
    println(aCard.simpleDescription())    //You'll see all the results in playground
}

enum Rank: Int {
    case Ace = 1
    case Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten
    case Jack, Queen, King

    func simpleDescription() -> String {
        switch self {
        case .Ace: return "ace"
        case .Jack: return "jack"
        case .Queen: return "queen"
        case .King: return "king"
        default: return String(self.toRaw())
        }
    }
}

enum Suit: Int {
    case Spades = 1
    case Hearts, Diamonds, Clubs

    func simpleDescription() -> String {
        switch self {
        case .Spades: return "spades"
        case .Hearts: return "hearts"
        case .Diamonds: return "diamonds"
        case .Clubs: return "clubs"
        }
    }

    func color() -> String {
        switch self {
        case .Spades, .Clubs: return "black"
        case .Hearts, .Diamonds: return "red"
        }
    }
}

struct Card {
    var rank: Rank
    var suit: Suit
    func simpleDescription() -> String {
        return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
    }

    static func createPokers() -> Card[] {
        let ranks = Array(Rank.Ace.toRaw()...Rank.King.toRaw())
        let suits = Array(Suit.Spades.toRaw()...Suit.Clubs.toRaw())
        let cards = suits.reduce(Card[]()) { (tempCards, suit) in
            tempCards + ranks.map { rank in
                Card(rank: Rank.fromRaw(rank)!, suit: Suit.fromRaw(suit)!)
            }
        }
        return cards
    }
}

实验内容是: 实验

在Card中添加一个方法，用于创建一副完整的牌，每一副牌都是rank和花色的组合。

因此，除了添加方法之外，没有修改或增强给定的代码(并且没有使用还没有教过的东西)，我想出了这个解决方案:

struct Card {
    var rank: Rank
    var suit: Suit

    func simpleDescription() -> String {
        return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
    }

    func createDeck() -> [Card] {
        var deck: [Card] = []
        for rank in Rank.Ace.rawValue...Rank.King.rawValue {
            for suit in Suit.Spades.rawValue...Suit.Clubs.rawValue {
                let card = Card(rank: Rank(rawValue: rank)!, suit: Suit(rawValue: suit)!)
                //println(card.simpleDescription())
                deck += [card]
            }
        }
        return deck
    }
}
let threeOfSpades = Card(rank: .Three, suit: .Spades)
let threeOfSpadesDescription = threeOfSpades.simpleDescription()
let deck = threeOfSpades.createDeck()