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

例如,我怎么做这样的事情:

for suit in Suit {
    // do something with suit
    print(suit.rawValue)
}

结果示例:

♠
♥
♦
♣

当前回答

Swift 4 + 2。

从Swift 4.2 (Xcode 10)开始,只需将协议一致性添加到CaseIterable中,就可以从allCases中受益。要添加这个协议一致性,你只需要在某个地方写:

extension Suit: CaseIterable {}

如果枚举是你自己的,你可以直接在声明中指定一致性:

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

然后下面的代码将打印所有可能的值:

Suit.allCases.forEach {
    print($0.rawValue)
}

与早期Swift版本的兼容性(3。X和4.x)

如果您需要支持Swift 3。x或4.0,你可以通过添加以下代码来模仿Swift 4.2的实现:

#if !swift(>=4.2)
public protocol CaseIterable {
    associatedtype AllCases: Collection where AllCases.Element == Self
    static var allCases: AllCases { get }
}
extension CaseIterable where Self: Hashable {
    static var allCases: [Self] {
        return [Self](AnySequence { () -> AnyIterator<Self> in
            var raw = 0
            var first: Self?
            return AnyIterator {
                let current = withUnsafeBytes(of: &raw) { $0.load(as: Self.self) }
                if raw == 0 {
                    first = current
                } else if current == first {
                    return nil
                }
                raw += 1
                return current
            }
        })
    }
}
#endif

其他回答

Xcode 10与Swift 4.2

enum Filter: String, CaseIterable {

    case salary = "Salary"
    case experience = "Experience"
    case technology = "Technology"
    case unutilized = "Unutilized"
    case unutilizedHV = "Unutilized High Value"

    static let allValues = Filter.allCases.map { $0.rawValue }
}

叫它

print(Filter.allValues)

打印:

[“薪酬”、“经验”、“技术”、“未利用”、“未利用的高价值”]


旧版本

对于表示Int的enum

enum Filter: Int {
    case salary
    case experience
    case technology
    case unutilized
    case unutilizedHV
    
    static let allRawValues = salary.rawValue...unutilizedHV.rawValue  // First to last case
    static let allValues = allRawValues.map { Filter(rawValue: $0)!.rawValue }
}

这样叫它:

print(Filter.allValues)

打印:

[0, 1, 2, 3, 4]


用于表示字符串的enum

enum Filter: Int {
    case salary
    case experience
    case technology
    case unutilized
    case unutilizedHV
    
    static let allRawValues = salary.rawValue...unutilizedHV.rawValue  // First to last case
    static let allValues = allRawValues.map { Filter(rawValue: $0)!.description }
}

extension Filter: CustomStringConvertible {
    var description: String {
        switch self {
        case .salary: return "Salary"
        case .experience: return "Experience"
        case .technology: return "Technology"
        case .unutilized: return "Unutilized"
        case .unutilizedHV: return "Unutilized High Value"
        }
    }
}

叫它

print(Filter.allValues)

打印:

[“薪酬”、“经验”、“技术”、“未利用”、“未利用的高价值”]

如果您仍然想为Rank和Suit使用枚举,这里有一个不那么神秘的例子。如果您想使用for-in循环遍历每个对象,只需将它们收集到一个Array中。

标准52张牌的例子:

enum Rank: Int {
    case Ace = 1, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King
    func name() -> 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 Diamonds = 1, Clubs, Hearts, Spades
    func name() -> String {
        switch self {
        case .Diamonds:
            return "diamonds"
        case .Clubs:
            return "clubs"
        case .Hearts:
            return "hearts"
        case .Spades:
            return "spades"
        default:
            return "NOT A VALID SUIT"
        }
    }
}

let Ranks = [
    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 Suits = [
    Suit.Diamonds,
    Suit.Clubs,
    Suit.Hearts,
    Suit.Spades
]


class Card {
    var rank: Rank
    var suit: Suit

    init(rank: Rank, suit: Suit) {
        self.rank = rank
        self.suit = suit
    }
}

class Deck {
    var cards = Card[]()

    init() {
        for rank in Ranks {
            for suit in Suits {
                cards.append(Card(rank: rank, suit: suit))
            }
        }
    }
}

var myDeck = Deck()
myDeck.cards.count  // => 52

以下是我的建议。这不是完全令人满意的(我对Swift和OOP很陌生!),但也许有人可以改进它。这个想法是让每个枚举提供自己的范围信息作为.first和.last属性。它只向每个枚举添加了两行代码:仍然有点硬编码,但至少它没有复制整个集合。它确实需要将Suit enum修改为Int类型,就像Rank enum一样,而不是无类型的。

而不是重复整个解决方案,下面是我添加到。在case语句之后的某个地方(Suit enum类似):

var first: Int { return Ace.toRaw() }
var last: Int { return King.toRaw() }

以及我用来将deck构建为String数组的循环。(问题定义没有说明牌组是如何构造的。)

func createDeck() -> [String] {
    var deck: [String] = []
    var card: String
    for r in Rank.Ace.first...Rank.Ace.last {
        for s in Suit.Hearts.first...Suit.Hearts.last {
            card = Rank.simpleDescription( Rank.fromRaw(r)!)() + " of " + Suit.simpleDescription( Suit.fromRaw(s)!)()
           deck.append( card)
       }
    }
    return deck
}

这并不令人满意,因为属性与元素而不是enum相关联。但它确实为“for”循环增加了清晰度。我希望它是Rank。而不是Rank.Ace.first。它适用于任何元素,但很难看。有人能演示一下如何将其提升到enum级别吗?

为了使它工作,我从Card结构中提取了createDeck方法。我不知道如何从该结构返回一个[String]数组,这似乎是一个糟糕的地方,把这样的方法无论如何。

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)
}

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
}