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()

在Swift 3中，当底层枚举有rawValue时，你可以实现Strideable协议。优点是不像其他建议那样创建值数组，并且标准的Swift“for in”循环工作，这是一个很好的语法。

// "Int" to get rawValue, and Strideable so we can iterate
enum MyColorEnum: Int, Strideable {
    case Red
    case Green
    case Blue
    case Black

    // required by Strideable
    typealias Stride = Int

    func advanced(by n:Stride) -> MyColorEnum {
        var next = self.rawValue + n
        if next > MyColorEnum.Black.rawValue {
            next = MyColorEnum.Black.rawValue
        }
        return MyColorEnum(rawValue: next)!
    }

    func distance(to other: MyColorEnum) -> Int {
        return other.rawValue - self.rawValue
    }

    // just for printing
    func simpleDescription() -> String {
        switch self {
        case .Red: return "Red"
        case .Green: return "Green"
        case .Blue: return "Blue"
        case .Black: return "Black"
        }
    }
}

// this is how you use it:
for i in MyColorEnum.Red ... MyColorEnum.Black {
    print("ENUM: \(i)")
}

与@Kametrixom的答案在这里，我相信返回一个数组将比返回AnySequence更好，因为你可以访问所有数组的好东西，如计数等。

以下是改写后的内容:

public protocol EnumCollection : Hashable {}
extension EnumCollection {
    public static func allValues() -> [Self] {
        typealias S = Self
        let retVal = 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
            }
        }

        return [S](retVal)
    }
}

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
}

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
}

其他的解决方法都是可行的，但它们都假设了可能的等级和花色的数量，或者第一和最后的等级是什么。的确，在可预见的未来，一副纸牌的布局可能不会有太大变化。然而，一般来说，编写尽可能少假设的代码会更简洁。我的解决方案:

我已经在Suit枚举中添加了一个原始类型，所以我可以使用Suit(rawValue:)来访问Suit案例:

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

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.rawValue)
        }
    }
}

在Card的createDeck()方法实现的下面。init(rawValue:)是一个可失败的初始化式，返回一个可选值。通过在两个while语句中展开并检查它的值，不需要假设Rank或Suit情况的数量:

struct Card {
    var rank: Rank
    var suit: Suit
    func simpleDescription() -> String {
        return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
    }
    func createDeck() -> [Card] {
        var n = 1
        var deck = [Card]()
        while let rank = Rank(rawValue: n) {
            var m = 1
            while let suit = Suit(rawValue: m) {
                deck.append(Card(rank: rank, suit: suit))
                m += 1
            }
            n += 1
        }
        return deck
    }
}

下面是如何调用createDeck方法:

let card = Card(rank: Rank.Ace, suit: Suit.Clubs)
let deck = card.createDeck()