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。

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
}

这个问题现在简单多了。以下是我的Swift 4.2解决方案:

enum Suit: Int, CaseIterable {
  case None
  case Spade, Heart, Diamond, Club

  static let allNonNullCases = Suit.allCases[Spade.rawValue...]
}

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

  static let allNonNullCases = Rank.allCases[Two.rawValue...]
}

func makeDeck(withJoker: Bool = false) -> [Card] {
  var deck = [Card]()
  for suit in Suit.allNonNullCases {
    for rank in Rank.allNonNullCases {
      deck.append(Card(suit: suit, rank: rank))
    }
  }
  if withJoker {
    deck.append(Card(suit: .None, rank: .Joker))
  }
  return deck
}

4。2:

我喜欢这个解决方案，我把找到“列表理解在Swift”。

它使用Int rawws而不是string，但它避免了键入两次，它允许自定义范围，并且不硬编码原始值。

这是我最初解决方案的Swift 4版本，但请参阅上面的4.2改进:

enum Suit: Int {
  case None
  case Spade, Heart, Diamond, Club

  static let allRawValues = Suit.Spade.rawValue...Suit.Club.rawValue
  static let allCases = Array(allRawValues.map{ Suit(rawValue: $0)! })
}
enum Rank: Int {
  case Joker
  case Two, Three, Four, Five, Six
  case Seven, Eight, Nine, Ten
  case Jack, Queen, King, Ace

  static let allRawValues = Rank.Two.rawValue...Rank.Ace.rawValue
  static let allCases = Array(allRawValues.map{ Rank(rawValue: $0)! })
}
func makeDeck(withJoker: Bool = false) -> [Card] {
  var deck = [Card]()
  for suit in Suit.allCases {
    for rank in Rank.allCases {
      deck.append(Card(suit: suit, rank: rank))
    }
  }
  if withJoker {
    deck.append(Card(suit: .None, rank: .Joker))
  }
  return deck
}

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()

在Swift 3中，当底层枚举有rawValue时，你可以实现Strideable协议。优点是不像其他建议那样创建值数组，并且标准的Swift“for in”循环工作，这是一个很好的语法。

// "Int" to get rawValue, and Strideable so we can iterate
enum MyColorEnum: Int, Strideable {
    case Red
    case Green
    case Blue
    case Black

    // required by Strideable
    typealias Stride = Int

    func advanced(by n:Stride) -> MyColorEnum {
        var next = self.rawValue + n
        if next > MyColorEnum.Black.rawValue {
            next = MyColorEnum.Black.rawValue
        }
        return MyColorEnum(rawValue: next)!
    }

    func distance(to other: MyColorEnum) -> Int {
        return other.rawValue - self.rawValue
    }

    // just for printing
    func simpleDescription() -> String {
        switch self {
        case .Red: return "Red"
        case .Green: return "Green"
        case .Blue: return "Blue"
        case .Black: return "Black"
        }
    }
}

// this is how you use it:
for i in MyColorEnum.Red ... MyColorEnum.Black {
    print("ENUM: \(i)")
}

它花了我一点，而不仅仅是一个方法在结构像swift书调用，但我在枚举中设置了下一个函数。我会使用一个协议，我不知道为什么，但有秩设置为int混乱。

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())
        }
    }
    mutating func next() -> Rank {
        var rank = self
        var rawrank = rank.toRaw()
        var nrank: Rank = self
        rawrank = rawrank + 1
        if let newRank = Rank.fromRaw(rawrank) {
            println("\(newRank.simpleDescription())")
            nrank = newRank
        } else {
            return self
        }
        return nrank
    }
}

enum Suit {
    case Spades, Hearts, Diamonds, Clubs
    func color() -> String {
        switch self {
        case .Spades, .Clubs:
            return "black"
        default:
            return "red"
        }
    }
    func simpleDescription() -> String {
        switch self {
        case .Spades:
            return "spades"
        case .Hearts:
            return "hearts"
        case .Diamonds:
            return "diamonds"
        case .Clubs:
            return "clubs"
        }
    }
    mutating func next() -> Suit {
        switch self {
        case .Spades:
            return Hearts
        case .Hearts:
            return Diamonds
        case .Diamonds:
            return Clubs
        case .Clubs:
            return Spades
        }
    }
}

struct Card {
    var rank: Rank
    var suit: Suit
    func deck() -> Card[] {
        var tRank = self.rank
        var tSuit = self.suit
        let tcards = 52 // we start from 0
        var cards: Card[] = []
        for i in 0..tcards {
            var card = Card(rank: tRank, suit: tSuit)
            cards.append(card)
            tRank = tRank.next()
            tSuit = tSuit.next()
        }
        return cards
    }
    func simpleDescription() -> String {
        return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
    }
}

var card = Card(rank: .Ace, suit: .Spades)
var deck = card.deck()

我使用了一些常识，但这可以通过将花色乘以等级来轻松纠正(如果你没有使用标准的桥牌，你必须相应地改变枚举，如果基本上只是通过不同的枚举进行步骤)。为了节省时间，我使用了ranks rawValues，如果你愿意，你也可以为西装做同样的事情。然而，这个例子没有它，所以我决定在不改变suit rawValue的情况下找出它