该解决方案在可读性和可维护性之间取得了适当的平衡。

struct Card {

    // ...

    static func deck() -> Card[] {
        var deck = Card[]()
        for rank in Rank.Ace.toRaw()...Rank.King.toRaw() {
            for suit in [Suit.Spades, .Hearts, .Clubs, .Diamonds] {
                let card = Card(rank: Rank.fromRaw(rank)!, suit: suit)
                deck.append(card)
            }
        }
    return deck
    }
}

let deck = Card.deck()

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++) {  
   ...  
}  

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
}

我在比特和字节中跌跌撞撞，创建了一个扩展，后来我发现它与@rintaro的答案非常相似。它是这样使用的:

enum E : EnumCollection {
    case A, B, C
}

Array(E.cases())    // [A, B, C]

值得注意的是，它可以在任何没有关联值的enum上使用。注意，这对于没有大小写的枚举不起作用。

与@rintaro的答案一样，这段代码使用枚举的底层表示。这种表示没有文档化，将来可能会改变，这会破坏它。我不建议在生产中使用这种方法。

代码(Swift 2.2, Xcode 7.3.1，不工作在Xcode 10):

protocol EnumCollection : Hashable {}
extension EnumCollection {
    static func cases() -> AnySequence<Self> {
        typealias S = Self
        return AnySequence { () -> AnyGenerator<S> in
            var raw = 0
            return AnyGenerator {
                let current : Self = withUnsafePointer(&raw) { UnsafePointer($0).memory }
                guard current.hashValue == raw else { return nil }
                raw += 1
                return current
            }
        }
    }
}

代码(Swift 3, Xcode 8.1，不工作在Xcode 10):

protocol EnumCollection : Hashable {}
extension EnumCollection {
    static func cases() -> AnySequence<Self> {
        typealias S = Self
        return AnySequence { () -> AnyIterator<S> in
            var raw = 0
            return AnyIterator {
                let current : Self = withUnsafePointer(to: &raw) { $0.withMemoryRebound(to: S.self, capacity: 1) { $0.pointee } }
                guard current.hashValue == raw else { return nil }
                raw += 1
                return current
            }
        }
    }
}

我不知道为什么我需要typealias，但编译器抱怨没有它。

另一个解决方案:

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

    static var count: Int {
        return 4   
    }

    init(index: Int) {
        switch index {
            case 0: self = .spades
            case 1: self = .hearts
            case 2: self = .diamonds
            default: self = .clubs
        }
    }
}

for i in 0..<Suit.count {
    print(Suit(index: i).rawValue)
}

enum Rank: Int
{
    case Ace = 0
    case Two, Three, Four, Five, Six, Seve, Eight, Nine, Ten
    case Jack, Queen, King
    case Count
}

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

struct Card
{
    var rank:Rank
    var suit:Suit
}

class Test
{
    func makeDeck() -> Card[]
    {
        let suitsCount:Int = Suit.Count.toRaw()
        let rankCount:Int = Rank.Count.toRaw()
        let repeatedCard:Card = Card(rank:Rank.Ace, suit:Suit.Spades)
        let deck:Card[] = Card[](count:suitsCount*rankCount, repeatedValue:repeatedCard)

        for i:Int in 0..rankCount
        {
            for j:Int in 0..suitsCount
            {
                deck[i*suitsCount+j] = Card(rank: Rank.fromRaw(i)!, suit: Suit.fromRaw(j)!)
            }
        }
        return deck
    }
}

根据Rick的回答:这要快5倍