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

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)
}

其他回答

这篇文章是相关的https://www.swift-studies.com/blog/2014/6/10/enumerating-enums-in-swift

基本上,提议的解决方案是

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

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

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

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()

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++) {  
   ...  
}  

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。

如果您仍然想为Rank和Suit使用枚举,这里有一个不那么神秘的例子。如果您想使用for-in循环遍历每个对象,只需将它们收集到一个Array中。

标准52张牌的例子:

enum Rank: Int {
    case Ace = 1, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King
    func name() -> 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 Diamonds = 1, Clubs, Hearts, Spades
    func name() -> String {
        switch self {
        case .Diamonds:
            return "diamonds"
        case .Clubs:
            return "clubs"
        case .Hearts:
            return "hearts"
        case .Spades:
            return "spades"
        default:
            return "NOT A VALID SUIT"
        }
    }
}

let Ranks = [
    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 Suits = [
    Suit.Diamonds,
    Suit.Clubs,
    Suit.Hearts,
    Suit.Spades
]


class Card {
    var rank: Rank
    var suit: Suit

    init(rank: Rank, suit: Suit) {
        self.rank = rank
        self.suit = suit
    }
}

class Deck {
    var cards = Card[]()

    init() {
        for rank in Ranks {
            for suit in Suits {
                cards.append(Card(rank: rank, suit: suit))
            }
        }
    }
}

var myDeck = Deck()
myDeck.cards.count  // => 52