如何枚举String类型的枚举？

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

例如，我该怎么做：

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

结果示例：

♠
♥
♦
♣

迅捷4.2+

从Swift 4.2（带有Xcode 10）开始，只需添加协议一致性CaseIterable即可从中受益allCases。要添加此协议一致性，您只需要在某处写：

extension Suit: CaseIterable {}

如果枚举是您自己的，则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，以下代码将打印所有可能的值：

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

与早期Swift版本（3.x和4.x）的兼容性

如果需要支持Swift 3.x或4.0，则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

这篇文章与这里有关https://www.swift-studies.com/blog/2014/6/10/enumerating-enums-in-swift

本质上，建议的解决方案是

enum ProductCategory : String {
     case Washers = "washers", Dryers = "dryers", Toasters = "toasters"

     static let allValues = [Washers, Dryers, Toasters]
}

for category in ProductCategory.allValues{
     //Do something
}

我制作了一个实用程序函数iterateEnum()来迭代任意enum类型的大小写。

这是示例用法：

enum Suit: String {
    case Spades = "♠"
    case Hearts = "♥"
    case Diamonds = "♦"
    case Clubs = "♣"
}

for f in iterateEnum(Suit) {
    println(f.rawValue)
}

哪个输出：

♠
♥
♦
♣

但是，这仅用于调试或测试目的：这依赖于多个未记录的Swift1.1编译器行为，因此，使用此方法后果自负。

这是代码：

func iterateEnum<T: Hashable>(_: T.Type) -> GeneratorOf<T> {
    var cast: (Int -> T)!
    switch sizeof(T) {
        case 0: return GeneratorOf(GeneratorOfOne(unsafeBitCast((), T.self)))
        case 1: cast = { unsafeBitCast(UInt8(truncatingBitPattern: $0), T.self) }
        case 2: cast = { unsafeBitCast(UInt16(truncatingBitPattern: $0), T.self) }
        case 4: cast = { unsafeBitCast(UInt32(truncatingBitPattern: $0), T.self) }
        case 8: cast = { unsafeBitCast(UInt64($0), T.self) }
        default: fatalError("cannot be here")
    }

    var i = 0
    return GeneratorOf {
        let next = cast(i)
        return next.hashValue == i++ ? next : nil
    }
}

基本思想是：

• 当案例的计数为时，的内存表示形式enum（不包括enum带有关联类型的s）只是案例的索引2...256，它与UInt8，when 257...65536，thes UInt16等等是相同的。因此，它可以unsafeBitcast来自相应的无符号整数类型。
• .hashValue 枚举值的和与案例的索引相同。
• .hashValue从无效索引进行位播的枚举值的总和为0。

已针对Swift2进行了修订，并从@Kametrixom的答案中实现了投放创意：

func iterateEnum<T: Hashable>(_: T.Type) -> AnyGenerator<T> {
    var i = 0
    return anyGenerator {
        let next = withUnsafePointer(&i) { UnsafePointer<T>($0).memory }
        return next.hashValue == i++ ? next : nil
    }
}

为Swift3修订：

func iterateEnum<T: Hashable>(_: T.Type) -> AnyIterator<T> {
    var i = 0
    return AnyIterator {
        let next = withUnsafePointer(to: &i) {
            $0.withMemoryRebound(to: T.self, capacity: 1) { $0.pointee }
        }
        if next.hashValue != i { return nil }
        i += 1
        return next
    }
}

为Swift3.0.1修订：

func iterateEnum<T: Hashable>(_: T.Type) -> AnyIterator<T> {
    var i = 0
    return AnyIterator {
        let next = withUnsafeBytes(of: &i) { $0.load(as: T.self) }
        if next.hashValue != i { return nil }
        i += 1
        return next
    }
}

其他解决方案也可行，但是它们都假设了例如可能的等级和西服的数量，或者首位和最后一位可能是多少。没错，在可预见的将来，一副纸牌的布局可能不会有太大变化。但是，总的来说，编写尽可能少的假设的代码更加整洁。我的解决方案：

我向Suit枚举添加了原始类型，因此可以Suit(rawValue:)用来访问Suit案例：

enum Suit: Int {
    case Spades = 1
    case Hearts, Diamonds, Clubs
    func simpleDescription() -> String {
        switch self {
            case .Spades:
                return "spades"
            case .Hearts:
                return "hearts"
            case .Diamonds:
                return "diamonds"
            case .Clubs:
                return "clubs"
        }
    }
    func color() -> String {
        switch self {
        case .Spades:
            return "black"
        case .Clubs:
            return "black"
        case .Diamonds:
            return "red"
        case .Hearts:
            return "red"
        }
    }
}

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.rawValue)
        }
    }
}

下面执行Card的createDeck()方法。init(rawValue:)是一个失败的初始化程序，并返回一个可选的。通过在两个while语句中解开并检查其值，无需假设Rank或的数量Suit：

struct Card {
    var rank: Rank
    var suit: Suit
    func simpleDescription() -> String {
        return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
    }
    func createDeck() -> [Card] {
        var n = 1
        var deck = [Card]()
        while let rank = Rank(rawValue: n) {
            var m = 1
            while let suit = Suit(rawValue: m) {
                deck.append(Card(rank: rank, suit: suit))
                m += 1
            }
            n += 1
        }
        return deck
    }
}

这是怎么称呼 createDeck方法的方法：

let card = Card(rank: Rank.Ace, suit: Suit.Clubs)
let deck = card.createDeck()

我偶然发现了位和字节，并创建了一个扩展，后来我发现它的扩展名与@rintaro的答案非常相似。它的用法如下：

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

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

值得注意的是，它可以在没有关联值的任何枚举中使用。请注意，这不适用于没有大小写的枚举。

与@rintaro的答案一样，此代码使用枚举的基本表示形式。此表示形式尚未记录，将来可能会更改，这会破坏它。 我不建议在生产中使用此方法。

代码（Swift 2.2，Xcode 7.3.1，不适用于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，不适用于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
            }
        }
    }
}

我不知道为什么需要typealias，但是编译器抱怨没有它。

您可以通过实现ForwardIndexType协议来遍历枚举。

该ForwardIndexType协议要求您定义一个successor()功能来逐步遍历元素。

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)!
}

在打开或关闭范围（..<或...）上进行迭代将在内部调用该successor()函数，该函数可让您编写以下代码：

// 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
}

现在，这个问题要容易得多。这是我的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之前的版本：

我喜欢在Swift中找到“ 列表理解 ”后将其组合在一起的解决方案。

它使用Int raws而不是Strings，但是它避免键入两次，它允许自定义范围，并且不对原始值进行硬编码。

这是我原始解决方案的Swift 4版本，但请参见上面的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
}

原则上，假设您在枚举的情况下不使用原始值分配，则可以这样做：

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())")
}

如果给枚举一个原始的Int值，它将使循环变得更加容易。

例如，您可以anyGenerator用来获取可以枚举所有值的生成器：

enum Suit: Int, CustomStringConvertible {
    case Spades, Hearts, Diamonds, Clubs
    var description: String {
        switch self {
        case .Spades:   return "Spades"
        case .Hearts:   return "Hearts"
        case .Diamonds: return "Diamonds"
        case .Clubs:    return "Clubs"
        }
    }
    static func enumerate() -> AnyGenerator<Suit> {
        var nextIndex = Spades.rawValue
        return anyGenerator { Suit(rawValue: nextIndex++) }
    }
}
// You can now use it like this:
for suit in Suit.enumerate() {
    suit.description
}
// or like this:
let allSuits: [Suit] = Array(Suit.enumerate())

但是，这似乎是一个相当普遍的模式，如果我们可以通过简单地遵循一个协议就可以使任何枚举类型成为可枚举的，那岂不是很好吗？有了Swift 2.0和协议扩展，现在我们可以了！

只需将其添加到您的项目中：

protocol EnumerableEnum {
    init?(rawValue: Int)
    static func firstValue() -> Int
}
extension EnumerableEnum {
    static func enumerate() -> AnyGenerator<Self> {
        var nextIndex = firstRawValue()
        return anyGenerator { Self(rawValue: nextIndex++) }
    }
    static func firstRawValue() -> Int { return 0 }
}

现在，无论何时创建枚举（只要它具有Int原始值），都可以通过遵守以下协议使其可枚举：

enum Rank: Int, EnumerableEnum {
    case Ace, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King
}
// ...
for rank in Rank.enumerate() { ... }

如果您的枚举值不是以0（默认）开头，请重写此firstRawValue方法：

enum DeckColor: Int, EnumerableEnum {
    case Red = 10, Blue, Black
    static func firstRawValue() -> Int { return Red.rawValue }
}
// ...
let colors = Array(DeckColor.enumerate())

最后西服类，包括更换simpleDescription与更标准CustomStringConvertible协议，将是这样的：

enum Suit: Int, CustomStringConvertible, EnumerableEnum {
    case Spades, Hearts, Diamonds, Clubs
    var description: String {
        switch self {
        case .Spades:   return "Spades"
        case .Hearts:   return "Hearts"
        case .Diamonds: return "Diamonds"
        case .Clubs:    return "Clubs"
        }
    }
}
// ...
for suit in Suit.enumerate() {
    print(suit.description)
}

Swift 3语法：

protocol EnumerableEnum {
    init?(rawValue: Int)
    static func firstRawValue() -> Int
}

extension EnumerableEnum {
    static func enumerate() -> AnyIterator<Self> {
        var nextIndex = firstRawValue()

        let iterator: AnyIterator<Self> = AnyIterator {
            defer { nextIndex = nextIndex + 1 }
            return Self(rawValue: nextIndex)
        }

        return iterator
    }

    static func firstRawValue() -> Int {
        return 0
    }
}

更新到Swift 2.2 +

func iterateEnum<T: Hashable>(_: T.Type) -> AnyGenerator<T> {
    var i = 0
    return AnyGenerator {
        let next = withUnsafePointer(&i) {
            UnsafePointer<T>($0).memory
        }
        if next.hashValue == i {
            i += 1
            return next
        } else {
            return nil
        }
    }
}

它已将代码更新为Swift 2.2形式@Kametrixom的答案

对于Swift 3.0+（非常感谢@Philip）

func iterateEnum<T: Hashable>(_: T.Type) -> AnyIterator<T> {
    var i = 0
    return AnyIterator {
        let next = withUnsafePointer(&i) {
            UnsafePointer<T>($0).pointee
        }
        if next.hashValue == i {
            i += 1
            return next
        } else {
            return nil
        }
    }
}

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)
}

我发现自己在.allValues整个代码中做很多事情。我终于找到了一种简单地遵循Iteratable协议并拥有rawValues()方法的方法。

protocol Iteratable {}
extension RawRepresentable where Self: RawRepresentable {

    static func iterateEnum<T: Hashable>(_: T.Type) -> AnyIterator<T> {
        var i = 0
        return AnyIterator {
            let next = withUnsafePointer(to: &i) {
                $0.withMemoryRebound(to: T.self, capacity: 1) { $0.pointee }
            }
            if next.hashValue != i { return nil }
            i += 1
            return next
        }
    }
}

extension Iteratable where Self: RawRepresentable, Self: Hashable {
    static func hashValues() -> AnyIterator<Self> {
        return iterateEnum(self)
    }

    static func rawValues() -> [Self.RawValue] {
        return hashValues().map({$0.rawValue})
    }
}


// Example
enum Grocery: String, Iteratable {
    case Kroger = "kroger"
    case HEB = "h.e.b."
    case Randalls = "randalls"
}

let groceryHashes = Grocery.hashValues() // AnyIterator<Grocery>
let groceryRawValues = Grocery.rawValues() // ["kroger", "h.e.b.", "randalls"]

带有Swift 4.2的Xcode 10

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 }
}

称它为

print(Filter.allValues)

印刷品：

[“工资”，“经验”，“技术”，“未利用”，“未利用的高价值”]

旧版本

对于enum代表Int

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 }
}

这样称呼它：

print(Filter.allValues)

印刷品：

[0，1，2，3，4]

对于enum代表String

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"
        }
    }
}

称它为

print(Filter.allValues)

印刷品：

[“工资”，“经验”，“技术”，“未利用”，“未利用的高价值”]

编辑： 快速发展提案 SE-0194枚举案例的派生集合提出了一个针对该问题的有针对性的解决方案。我们在Swift 4.2及更高版本中看到了它。该提案还指出了一些变通办法，这些变通办法与此处已经提到的变通办法相似，但是仍然有可能引起人们的兴趣。

为了完整起见，我还将保留原始帖子。

这是基于@Peymmankh的答案的另一种方法，适用于Swift 3。

public protocol EnumCollection: Hashable {}

extension EnumCollection {

public static func allValues() -> [Self] {
    typealias S = Self

    let retVal = AnySequence { () -> AnyIterator<S> in
        var raw = 0
        return AnyIterator {
            let current = withUnsafePointer(to: &raw) {
                 $0.withMemoryRebound(to: S.self, capacity: 1) { $0.pointee }
            }
            guard current.hashValue == raw else { return nil }
            raw += 1
            return current
        }
    }

    return [S](retVal)
}

enum Rank: Int {
    ...
    static let ranks = (Rank.Ace.rawValue ... Rank.King.rawValue).map{Rank(rawValue: $0)! }

}
enum Suit {
    ...
    static let suits = [Spades, Hearts, Diamonds, Clubs]
}

struct Card {
    ...
    static func fullDesk() -> [Card] {
        var desk: [Card] = []
        for suit in Suit.suits {
            for rank in Rank.ranks {
                desk.append(Card(rank: rank,suit: suit))
            }
        }
        return desk
    }
}

这个怎么样？

您可以尝试这样枚举

enum Planet: String {
    case Mercury
    case Venus
    case Earth
    case Mars

    static var enumerate: [Planet] {
        var a: [Planet] = []
        switch Planet.Mercury {
            case .Mercury: a.append(.Mercury); fallthrough
            case .Venus: a.append(.Venus); fallthrough
            case .Earth: a.append(.Earth); fallthrough
            case .Mars: a.append(.Mars)
        }
    return a
    }
}

Planet.enumerate // [Mercury, Venus, Earth, Mars]

在Swift 3中，当基础枚举具有时rawValue，您可以实现Strideable协议。优点是不会像其他建议中那样创建值数组，并且标准Swift“ for in”循环也可以工作，这使语法很不错。

// "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)")
}

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()

抱歉，我的答案特定于我在需要做的事情中使用这篇文章的方式。对于那些偶然发现此问题的人，寻找一种在枚举中找到案例的方法，这是解决问题的方法（Swift 2中的新增功能）：

编辑：小写的camelCase现在是Swift 3枚举值的标准

// From apple docs: If the raw-value type is specified as String and you don’t assign values to the cases explicitly, each unassigned case is implicitly assigned a string with the same text as the name of that case.

enum Theme: String
    {
    case white, blue, green, lavender, grey
    }

func loadTheme(theme: String)
    {
    // this checks the string against the raw value of each enum case (note that the check could result in a nil value, since it's an optional, which is why we introduce the if/let block
    if let testTheme = Theme(rawValue: theme)
        {
        // testTheme is guaranteed to have an enum value at this point
        self.someOtherFunction(testTheme)
        }
    }

对于那些想枚举枚举的人，此页面上给出的答案是正确的，该答案包括一个包含所有枚举值数组的静态var / let。Apple最新的tvOS示例代码包含此完全相同的技术。

话虽这么说，他们应该在语言中建立更方便的机制（Apple，您在听吗？）！

实验是：实验

向Card添加一种方法，以创建一副完整的纸牌，每个等级和花色组合都包含一张纸牌。

因此，除了添加方法（没有使用尚未讲授的内容）外，没有修改或增强给定的代码，我想出了以下解决方案：

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()

这是我用来迭代enum和从一个提供多个值类型的方法enum

enum IterateEnum: Int {
    case Zero
    case One
    case Two
    case Three
    case Four
    case Five
    case Six
    case Seven

    //tuple allows multiple values to be derived from the enum case, and
    //since it is using a switch with no default, if a new case is added,
    //a compiler error will be returned if it doesn't have a value tuple set
    var value: (french: String, spanish: String, japanese: String) {
        switch self {
        case .Zero: return (french: "zéro", spanish: "cero", japanese: "nuru")
        case .One: return (french: "un", spanish: "uno", japanese: "ichi")
        case .Two: return (french: "deux", spanish: "dos", japanese: "ni")
        case .Three: return (french: "trois", spanish: "tres", japanese: "san")
        case .Four: return (french: "quatre", spanish: "cuatro", japanese: "shi")
        case .Five: return (french: "cinq", spanish: "cinco", japanese: "go")
        case .Six: return (french: "six", spanish: "seis", japanese: "roku")
        case .Seven: return (french: "sept", spanish: "siete", japanese: "shichi")
        }
    }

    //Used to iterate enum or otherwise access enum case by index order.
    //Iterate by looping until it returns nil
    static func item(index: Int) -> IterateEnum? {
        return IterateEnum.init(rawValue: index)
    }

    static func numberFromSpanish(number: String) -> IterateEnum? {
        return findItem { $0.value.spanish == number }
    }

    //use block to test value property to retrieve the enum case        
    static func findItem(predicate: ((_: IterateEnum) -> Bool)) -> IterateEnum? {

        var enumIndex: Int = -1
        var enumCase: IterateEnum?

        //Iterate until item returns nil
        repeat {
            enumIndex += 1
            enumCase = IterateEnum.item(index: enumIndex)

            if let eCase = enumCase {

                if predicate(eCase) {
                    return eCase
                }
            }
        } while enumCase != nil
        return nil
    }
}

var enumIndex: Int = -1
var enumCase: IterateEnum?

// Iterate until item returns nil
repeat {
    enumIndex += 1
    enumCase = IterateEnum.item(index: enumIndex)
    if let eCase = enumCase {
        print("The number \(eCase) in french: \(eCase.value.french), spanish: \(eCase.value.spanish), japanese: \(eCase.value.japanese)")
    }
} while enumCase != nil

print("Total of \(enumIndex) cases")

let number = IterateEnum.numberFromSpanish(number: "siete")

print("siete in japanese: \((number?.value.japanese ?? "Unknown"))")

这是输出：

在法国的数字零：零，西班牙语：CERO，日本：努鲁
头号法语：联合国，西班牙语：UNO，日本：ICHI
二号法文：德塞夫勒，西班牙语：DOS，日本：NI
在法国排名第三：trois，西班牙语：tres，日语：san
法语中的第四位：quatre，西班牙语：cuatro，日语：shi
法语中的第五位：cinq，西班牙语：cinco，日语：go
法语中的第六位：六个，西班牙语： seis，日语：roku
法语中的第七：sept，西班牙语：siete，日语：shichi

共有8例

siete日语：shichi

更新

我最近创建了一个协议来处理枚举。该协议需要一个具有Int原始值的枚举：

protocol EnumIteration {

    //Used to iterate enum or otherwise access enum case by index order. Iterate by looping until it returns nil

    static func item(index:Int) -> Self?
    static func iterate(item:((index:Int, enumCase:Self)->()), completion:(()->())?) {
    static func findItem(predicate:((enumCase:Self)->Bool)) -> Self?
    static func count() -> Int
}

extension EnumIteration where Self: RawRepresentable, Self.RawValue == Int {

    //Used to iterate enum or otherwise access enum case by index order. Iterate by looping until it returns nil
    static func item(index:Int) -> Self? {
        return Self.init(rawValue: index)
    }

    static func iterate(item:((index:Int, enumCase:Self)->()), completion:(()->())?) {

        var enumIndex:Int = -1
        var enumCase:Self?

        //Iterate until item returns nil
        repeat {
            enumIndex += 1
            enumCase = Self.item(enumIndex)

            if let eCase = enumCase {
                item(index: enumIndex, enumCase: eCase)
            }
        } while enumCase != nil
        completion?()
    }

    static func findItem(predicate:((enumCase:Self)->Bool)) -> Self? {

        var enumIndex:Int = -1
        var enumCase:Self?

        //Iterate until item returns nil
        repeat {
            enumIndex += 1
            enumCase = Self.item(enumIndex)

            if let eCase = enumCase {

                if predicate(enumCase:eCase) {
                    return eCase
                }
            }
        } while enumCase != nil
        return nil
    }

    static func count() -> Int {
        var enumIndex:Int = -1
        var enumCase:Self?

        //Iterate until item returns nil
        repeat {
            enumIndex += 1
            enumCase = Self.item(enumIndex)
        } while enumCase != nil

        //last enumIndex (when enumCase == nil) is equal to the enum count
        return enumIndex
    }
}

这似乎是一种hack，但是如果您使用原始值，则可以执行以下操作

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

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

Swift 2.0我的建议是在这里处理：

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
    }
}

与@Kametrixom一样在这里回答我相信返回数组比返回AnySequence更好，因为您可以访问Array的所有功能，例如count等。

这是重写：

public protocol EnumCollection : Hashable {}
extension EnumCollection {
    public static func allValues() -> [Self] {
        typealias S = Self
        let retVal = 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
            }
        }

        return [S](retVal)
    }
}

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)
}

这是来自Swift 2.0的相当古老的文章。现在，这里有一些更好的解决方案使用了Swift 3.0的新功能： 在Swift 3.0中遍历枚举

在这个问题上，有一个解决方案使用了一项新功能（在我撰写此编辑时尚未发布）Swift 4.2： 如何获得Swift枚举的计数？

这个线程中有很多好的解决方案，而其他一些却非常复杂。我喜欢尽可能简化。这是一个可能会或可能无法满足不同需求的解决方案，但我认为它在大多数情况下都能很好地工作：

enum Number: String {
    case One
    case Two
    case Three
    case Four
    case EndIndex

    func nextCase () -> Number
    {
        switch self {
        case .One:
            return .Two
        case .Two:
            return .Three
        case .Three:
            return .Four
        case .Four:
            return .EndIndex

        /* 
        Add all additional cases above
        */
        case .EndIndex:
            return .EndIndex
        }
    }

    static var allValues: [String] {
        var array: [String] = Array()
        var number = Number.One

        while number != Number.EndIndex {
            array.append(number.rawValue)
            number = number.nextCase()
        }
        return array
    }
}

要迭代：

for item in Number.allValues {
    print("number is: \(item)")
}

枚举有toRaw()和fromRaw()方法。因此，如果原始值为an Int，则可以从头到尾进行迭代enum：

enum Suit: Int {
    case Spades = 1
    case Hearts, Diamonds, Clubs
    func simpleDescription() -> String {
        switch self {
        case .Spades:
            return "spades"
        case .Hearts:
            return "hearts"
        case .Diamonds:
            return "diamonds"
        case .Clubs:
            return "clubs"
        }
    }
}

for i in Suit.Spades.toRaw()...Suit.Clubs.toRaw() {
    if let covertedSuit = Suit.fromRaw(i) {
        let description = covertedSuit.simpleDescription()
    }
}

一个陷阱是您需要在运行该simpleDescription方法之前测试可选值，因此我们convertedSuit首先将其设置为值，然后将一个常量设置为convertedSuit.simpleDescription()

这是我建议的方法。这并不完全令人满意（我对Swift和OOP还是很陌生！），但也许有人可以对其进行完善。想法是让每个枚举提供其自身的范围信息.first和.last属性。它向每个枚举只添加了两行代码：仍然有些硬编码，但至少没有重复整个代码集。它确实需要Suit像Rank枚举一样将枚举修改为一个Int ，而不是未键入。

这不是在echo整个解决方案，而是.在case语句后的某个位置（Suit枚举类似）添加到枚举的代码：

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

和我用来将牌组构建为String数组的循环。（问题定义未说明平台的结构。）

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
}