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

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
}

其他回答

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。

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
}

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
}

原则上,如果你不为enum的大小写使用原始值赋值,这样做是可能的:

enum RankEnum: Int {
  case Ace
  case One
  case Two
}

class RankEnumGenerator: Generator {
    var i = 0
    typealias Element = RankEnum
    func next() -> Element? {
        let r = RankEnum.fromRaw(i)
        i += 1
        return r
    }
}

extension RankEnum {
    static func enumerate() -> SequenceOf<RankEnum> {
        return SequenceOf<RankEnum>({ RankEnumGenerator() })
    }
}

for r in RankEnum.enumerate() {
    println("\(r.toRaw())")
}