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。

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())
        }
    }
}

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, .Clubs: return "black"
        case .Hearts, .Diamonds: return "red"
        }
    }
}

struct Card {
    var rank: Rank
    var suit: Suit
    func simpleDescription() -> String {
        return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
    }

    static func createPokers() -> Card[] {
        let ranks = Array(Rank.Ace.toRaw()...Rank.King.toRaw())
        let suits = Array(Suit.Spades.toRaw()...Suit.Clubs.toRaw())
        let cards = suits.reduce(Card[]()) { (tempCards, suit) in
            tempCards + ranks.map { rank in
                Card(rank: Rank.fromRaw(rank)!, suit: Suit.fromRaw(suit)!)
            }
        }
        return cards
    }
}

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

(改进Karthik Kumar的回答)

这个解决方案是使用编译器来保证你不会错过一个case。

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

    static var enumerate: [Suit] {
        switch Suit.spades {
        // make sure the two lines are identical ^_^
        case        .spades, .hearts, .diamonds, .clubs:
            return [.spades, .hearts, .diamonds, .clubs]
        }
    }
}

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)
}

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
}