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"]

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
}

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
}

这是一个相当老的帖子，来自Swift 2.0。现在有一些更好的解决方案，使用了swift 3.0的新特性: 在Swift 3.0中迭代一个Enum

关于这个问题，有一个解决方案，它使用了Swift 4.2的一个新功能(在我写这篇编辑时还没有发布): 我如何得到一个Swift枚举的计数?

在这个帖子中有很多好的解决方案，但其中一些非常复杂。我喜欢尽可能地简化。这里有一个解决方案，可能适用于不同的需求，但我认为它在大多数情况下都很好:

enum Number: String {
    case One
    case Two
    case Three
    case Four
    case EndIndex

    func nextCase () -> Number
    {
        switch self {
        case .One:
            return .Two
        case .Two:
            return .Three
        case .Three:
            return .Four
        case .Four:
            return .EndIndex

        /* 
        Add all additional cases above
        */
        case .EndIndex:
            return .EndIndex
        }
    }

    static var allValues: [String] {
        var array: [String] = Array()
        var number = Number.One

        while number != Number.EndIndex {
            array.append(number.rawValue)
            number = number.nextCase()
        }
        return array
    }
}

迭代:

for item in Number.allValues {
    print("number is: \(item)")
}

我发现了一种有点俗气但更安全的方法，它不需要键入两次值或引用枚举值的内存，因此不太可能损坏。

基本上，与其使用枚举，不如创建一个具有单个实例的结构体，并将所有enum-values设置为常量。然后可以使用Mirror查询变量

public struct Suit{

    // the values
    let spades = "♠"
    let hearts = "♥"
    let diamonds = "♦"
    let clubs = "♣"

    // make a single instance of the Suit struct, Suit.instance
    struct SStruct{static var instance: Suit = Suit()}
    static var instance : Suit{
        get{return SStruct.instance}
        set{SStruct.instance = newValue}
    }

    // an array with all of the raw values
    static var allValues: [String]{
        var values = [String]()

        let mirror = Mirror(reflecting: Suit.instance)
        for (_, v) in mirror.children{
            guard let suit = v as? String else{continue}
            values.append(suit)
        }

        return values
    }
}

如果使用此方法，则需要使用Suit.instance.clubs或Suit.instance.spades来获取单个值

但所有这些都太无聊了……让我们做一些事情，使它更像一个真正的enum!

public struct SuitType{

    // store multiple things for each suit
    let spades = Suit("♠", order: 4)
    let hearts = Suit("♥", order: 3)
    let diamonds = Suit("♦", order: 2)
    let clubs = Suit("♣", order: 1)

    struct SStruct{static var instance: SuitType = SuitType()}
    static var instance : SuitType{
        get{return SStruct.instance}
        set{SStruct.instance = newValue}
    }

    // a dictionary mapping the raw values to the values
    static var allValuesDictionary: [String : Suit]{
        var values = [String : Suit]()

        let mirror = Mirror(reflecting: SuitType.instance)
        for (_, v) in mirror.children{
            guard let suit = v as? Suit else{continue}
            values[suit.rawValue] = suit
        }

        return values
    }
}

public struct Suit: RawRepresentable, Hashable{
    public var rawValue: String
    public typealias RawValue = String

    public var hashValue: Int{
        // find some integer that can be used to uniquely identify
        // each value. In this case, we could have used the order
        // variable because it is a unique value, yet to make this
        // apply to more cases, the hash table address of rawValue
        // will be returned, which should work in almost all cases
        // 
        // you could also add a hashValue parameter to init() and
        // give each suit a different hash value
        return rawValue.hash
    }

    public var order: Int
    public init(_ value: String, order: Int){
        self.rawValue = value
        self.order = order
    }

    // an array of all of the Suit values
    static var allValues: [Suit]{
        var values = [Suit]()

        let mirror = Mirror(reflecting: SuitType.instance)
        for (_, v) in mirror.children{
            guard let suit = v as? Suit else{continue}
            values.append(suit)
        }

        return values
    }

    // allows for using Suit(rawValue: "♦"), like a normal enum
    public init?(rawValue: String){
        // get the Suit from allValuesDictionary in SuitType, or return nil if that raw value doesn't exist
        guard let suit = SuitType.allValuesDictionary[rawValue] else{return nil}
        // initialize a new Suit with the same properties as that with the same raw value
        self.init(suit.rawValue, order: suit.order)
    }
}

你现在可以做

let allSuits: [Suit] = Suit.allValues

or

for suit in Suit.allValues{
   print("The suit \(suit.rawValue) has the order \(suit.order)")
}

然而，要获得一个单一，你仍然需要使用SuitType.instance.spades或SuitType.instance.hearts。为了更加直观，您可以向Suit添加一些允许您使用Suit.type的代码。*而不是SuitType.instance.*

public struct Suit: RawRepresentable, Hashable{
   // ...your code...

   static var type = SuitType.instance

   // ...more of your code...
}

您现在可以使用Suit.type.diamonds而不是SuitType.instance。diamonds，或者Suit.type.clubs而不是SuitType.instance.clubs

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()