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

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?

其他回答

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)
        }
    }

    //...
}

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
}

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()
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?

更新代码:Swift 4.2/Swift 5

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

按问题访问输出:

for suitKey in Suit.allCases {
    print(suitKey.rawValue)
}

输出:

♠
♥
♦
♣

CaseIterable:提供其所有值的集合。 符合CaseIterable协议的类型通常是没有关联值的枚举。当使用CaseIterable类型时,您可以通过使用该类型的allCases属性访问该类型的所有案例的集合。

对于访问case,我们使用。allcases。更多信息请点击https://developer.apple.com/documentation/swift/caseiterable