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

枚举有toRaw()和fromRaw()方法。所以如果你的原始值是Int,你可以从第一个枚举迭代到最后一个枚举:

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()
    }
}

一个问题是在运行simpleDescription方法之前需要测试可选值,因此我们首先将convertedSuit设置为我们的值,然后将常量设置为convertedSuit.simpleDescription()

其他回答

下面是我用来迭代枚举和从一个枚举提供多个值类型的方法

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"))")

输出如下:

法语中的数字Zero: zéro,西班牙语中的数字cero,日语中的数字nuru 数字一在法语中是un,西班牙语中是uno,日语中是ichi 法语中的数字2是deux,西班牙语中的数字2是dos,日语中的数字2是ni 法语中的“三”是“trois”,西班牙语中的“tres”,日语中的“san” 法语中的“四”是quatre,西班牙语中的“四”是cuatro,日语中的“四”是shi 数字五在法语中是cinq,西班牙语中是cinco,日语中是go 数字6在法语中是Six,西班牙语是seis,日语是roku 法语中的数字“七”是“sept”,西班牙语中的“siete”,日语中的“shichi”

共8例

Siete在日语中的意思是: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Int原始值的enum:

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
    }
}

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)
        }
    }

    //...
}
enum Rank: Int
{
    case Ace = 0
    case Two, Three, Four, Five, Six, Seve, Eight, Nine, Ten
    case Jack, Queen, King
    case Count
}

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

struct Card
{
    var rank:Rank
    var suit:Suit
}

class Test
{
    func makeDeck() -> Card[]
    {
        let suitsCount:Int = Suit.Count.toRaw()
        let rankCount:Int = Rank.Count.toRaw()
        let repeatedCard:Card = Card(rank:Rank.Ace, suit:Suit.Spades)
        let deck:Card[] = Card[](count:suitsCount*rankCount, repeatedValue:repeatedCard)

        for i:Int in 0..rankCount
        {
            for j:Int in 0..suitsCount
            {
                deck[i*suitsCount+j] = Card(rank: Rank.fromRaw(i)!, suit: Suit.fromRaw(j)!)
            }
        }
        return deck
    }
}

根据Rick的回答:这要快5倍

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。

实验内容是: 实验

在Card中添加一个方法,用于创建一副完整的牌,每一副牌都是rank和花色的组合。

因此,除了添加方法之外,没有修改或增强给定的代码(并且没有使用还没有教过的东西),我想出了这个解决方案:

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()