在处理Swift 2.0时，以下是我的建议:

我已经将原始类型添加到Suit enum

enum Suit: Int {

然后:

struct Card {
    var rank: Rank
    var suit: Suit


    func fullDeck()-> [Card] {

        var deck = [Card]()

        for i in Rank.Ace.rawValue...Rank.King.rawValue {

            for j in Suit.Spades.rawValue...Suit.Clubs.rawValue {

                deck.append(Card(rank:Rank(rawValue: i)! , suit: Suit(rawValue: j)!))
            }
        }

        return deck
    }
}

您可以通过实现ForwardIndexType协议来迭代枚举。

ForwardIndexType协议要求您定义一个继任者()函数来逐级遍历元素。

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

    // ... other functions

    // Option 1 - Figure it out by hand
    func successor() -> Rank {
        switch self {
            case .Ace:
              return .Two
            case .Two:
              return .Three

            // ... etc.

            default:
              return .King
        }
    }

    // Option 2 - Define an operator!
    func successor() -> Rank {
        return self + 1
    }
}

// NOTE: The operator is defined OUTSIDE the class
func + (left: Rank, right: Int) -> Rank {
    // I'm using to/from raw here, but again, you can use a case statement
    // or whatever else you can think of

    return left == .King ? .King : Rank(rawValue: left.rawValue + right)!
}

在开或闭范围内迭代(..<或…)将在内部调用继任者()函数，允许你这样写:

// Under the covers, successor(Rank.King) and successor(Rank.Ace) are called to establish limits
for r in Rank.Ace...Rank.King {
    // Do something useful
}

这看起来像一个黑客，但如果你使用原始值，你可以这样做

enum Suit: Int {  
    case Spades = 0, Hearts, Diamonds, Clubs  
 ...  
}  

var suitIndex = 0  
while var suit = Suit.fromRaw(suitIndex++) {  
   ...  
}  

我使用计算属性，它返回所有值的数组(感谢这篇文章http://natecook.com/blog/2014/10/loopy-random-enum-ideas/)。但是，它也使用int原始值，但我不需要在单独的属性中重复枚举的所有成员。

Xcode 6.1在如何使用rawValue获取enum成员方面做了一点改变，所以我修正了listing。还修复了第一个rawValue错误的小错误。

enum ValidSuits: Int {
    case Clubs = 0, Spades, Hearts, Diamonds
    func description() -> String {
        switch self {
        case .Clubs:
            return "♣︎"
        case .Spades:
            return "♠︎"
        case .Diamonds:
            return "♦︎"
        case .Hearts:
            return "♥︎"
        }
    }

    static var allSuits: [ValidSuits] {
        return Array(
            SequenceOf {
                () -> GeneratorOf<ValidSuits> in
                var i=0
                return GeneratorOf<ValidSuits> {
                    return ValidSuits(rawValue: i++)
                }
            }
        )
    }
}

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()

我使用了下面的方法，假设我知道哪个是Rank enum中的最后一个值，所有的Rank在Ace之后都有增量值

我喜欢这种方式，因为它干净，小，容易理解

 func cardDeck() -> Card[] {
     var cards: Card[] = []
     let minRank = Rank.Ace.toRaw()
     let maxRank = Rank.King.toRaw()

     for rank in minRank...maxRank {
         if var convertedRank: Rank = Rank.fromRaw(rank) {
             cards.append(Card(rank: convertedRank, suite: Suite.Clubs))
             cards.append(Card(rank: convertedRank, suite: Suite.Diamonds))
             cards.append(Card(rank: convertedRank, suite: Suite.Hearts))
             cards.append(Card(rank: convertedRank, suite: Suite.Spades))
         }
    }

    return cards
}