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

我发现自己在代码中经常使用. allvalues。我终于找到了一种方法来简单地遵循Iteratable协议并拥有一个rawValues()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"]

其他回答

我创建了一个实用函数iterateEnum(),用于迭代任意枚举类型的情况。

下面是示例用法:

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

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

输出:

♠
♥
♦
♣

但是,这仅用于调试或测试目的:这依赖于几个未记录的Swift1.1编译器行为,因此,使用它的风险由您自己承担。

代码如下:

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
    }
}

其基本思想是:

枚举的内存表示,不包括有关联类型的枚举,只是一个案例的索引,当案例的计数是2…256,它和UInt8是一样的,当257…65536,它是UInt16等等。因此,它可以是unsafeBitcast对应的无符号整数类型。 枚举值的. hashvalue与case的索引相同。 从无效索引位转换的枚举值的. hashvalue为0。


为Swift2修改,并从@Kametrixom的回答中实现了选角想法:

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
    }
}

对Swift3的修订:

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
    }
}

针对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
    }
}

有时,您可能会处理具有底层原始整数类型的枚举类型,这种类型在整个软件开发生命周期中都会发生变化。下面是一个很适合这种情况的例子:

public class MyClassThatLoadsTexturesEtc
{
    //...

    // Colors used for gems and sectors.
    public enum Color: Int
    {
        // Colors arranged in order of the spectrum.
        case First = 0
        case Red, Orange, Yellow, Green, Blue, Purple, Pink
        // --> Add more colors here, between the first and last markers.
        case Last
    }

    //...

    public func preloadGems()
    {
        // Preload all gems.
        for i in (Color.First.toRaw() + 1) ..< (Color.Last.toRaw())
        {
            let color = Color.fromRaw(i)!
            loadColoredTextures(forKey: color)
        }
    }

    //...
}

在Swift中,枚举类型可以像EnumType一样访问。案例:

let tableView = UITableView(frame: self.view. view)UITableViewStyle.Plain)

大多数情况下,只有当您有几个选项可以使用,并且确切地知道在每个选项上要做什么时,才会使用枚举类型。

在处理枚举类型时,使用for-in结构没有太大意义。

你可以这样做,例如:

func sumNumbers(numbers : Int...) -> Int {
    var sum = 0

    for number in numbers{
        sum += number
    }

    return sum
}

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

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