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

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
}

其他回答

对不起,我的回答是具体到我如何在我需要做的事情中使用这篇文章。对于那些无意中遇到这个问题的人,寻找一种方法在枚举中找到一个case,这是一种方法(Swift 2新增):

编辑:小写驼峰现在是Swift 3 enum值的标准

// From apple docs: If the raw-value type is specified as String and you don’t assign values to the cases explicitly, each unassigned case is implicitly assigned a string with the same text as the name of that case.

enum Theme: String
    {
    case white, blue, green, lavender, grey
    }

func loadTheme(theme: String)
    {
    // this checks the string against the raw value of each enum case (note that the check could result in a nil value, since it's an optional, which is why we introduce the if/let block
    if let testTheme = Theme(rawValue: theme)
        {
        // testTheme is guaranteed to have an enum value at this point
        self.someOtherFunction(testTheme)
        }
    }

对于那些对枚举感到疑惑的人来说,本页上给出的答案包括一个包含所有枚举值的数组的静态var/let是正确的。最新的苹果tvOS示例代码包含了完全相同的技术。

也就是说,他们应该在语言中构建一个更方便的机制(苹果,你在听吗?)

我使用计算属性,它返回所有值的数组(感谢这篇文章http://natecook.com/blog/2014/10/loopy-random-enum-ideas/)。但是,它也使用int原始值,但我不需要在单独的属性中重复枚举的所有成员。

Xcode 6.1在如何使用rawValue获取enum成员方面做了一点改变,所以我修正了listing。还修复了第一个rawValue错误的小错误。

enum ValidSuits: Int {
    case Clubs = 0, Spades, Hearts, Diamonds
    func description() -> String {
        switch self {
        case .Clubs:
            return "♣︎"
        case .Spades:
            return "♠︎"
        case .Diamonds:
            return "♦︎"
        case .Hearts:
            return "♥︎"
        }
    }

    static var allSuits: [ValidSuits] {
        return Array(
            SequenceOf {
                () -> GeneratorOf<ValidSuits> in
                var i=0
                return GeneratorOf<ValidSuits> {
                    return ValidSuits(rawValue: i++)
                }
            }
        )
    }
}

如果你给枚举一个原始的Int值,它将使循环更容易。

例如,你可以使用anyGenerator来获得一个可以枚举你的值的生成器:

enum Suit: Int, CustomStringConvertible {
    case Spades, Hearts, Diamonds, Clubs
    var description: String {
        switch self {
        case .Spades:   return "Spades"
        case .Hearts:   return "Hearts"
        case .Diamonds: return "Diamonds"
        case .Clubs:    return "Clubs"
        }
    }
    static func enumerate() -> AnyGenerator<Suit> {
        var nextIndex = Spades.rawValue
        return anyGenerator { Suit(rawValue: nextIndex++) }
    }
}
// You can now use it like this:
for suit in Suit.enumerate() {
    suit.description
}
// or like this:
let allSuits: [Suit] = Array(Suit.enumerate())

然而,这看起来像一个相当常见的模式,如果我们可以通过简单地遵循协议使任何枚举类型可枚举,这不是很好吗?有了Swift 2.0和协议扩展,现在我们可以了!

简单地添加到你的项目:

protocol EnumerableEnum {
    init?(rawValue: Int)
    static func firstValue() -> Int
}
extension EnumerableEnum {
    static func enumerate() -> AnyGenerator<Self> {
        var nextIndex = firstRawValue()
        return anyGenerator { Self(rawValue: nextIndex++) }
    }
    static func firstRawValue() -> Int { return 0 }
}

现在,任何时候你创建一个枚举(只要它有一个Int原始值),你可以通过遵循协议使它可枚举:

enum Rank: Int, EnumerableEnum {
    case Ace, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King
}
// ...
for rank in Rank.enumerate() { ... }

如果你的枚举值不以0开头(默认值),重写firstRawValue方法:

enum DeckColor: Int, EnumerableEnum {
    case Red = 10, Blue, Black
    static func firstRawValue() -> Int { return Red.rawValue }
}
// ...
let colors = Array(DeckColor.enumerate())

最后一个Suit类,包括用更标准的CustomStringConvertible协议替换simpleDescription,看起来像这样:

enum Suit: Int, CustomStringConvertible, EnumerableEnum {
    case Spades, Hearts, Diamonds, Clubs
    var description: String {
        switch self {
        case .Spades:   return "Spades"
        case .Hearts:   return "Hearts"
        case .Diamonds: return "Diamonds"
        case .Clubs:    return "Clubs"
        }
    }
}
// ...
for suit in Suit.enumerate() {
    print(suit.description)
}

Swift 3语法:

protocol EnumerableEnum {
    init?(rawValue: Int)
    static func firstRawValue() -> Int
}

extension EnumerableEnum {
    static func enumerate() -> AnyIterator<Self> {
        var nextIndex = firstRawValue()

        let iterator: AnyIterator<Self> = AnyIterator {
            defer { nextIndex = nextIndex + 1 }
            return Self(rawValue: nextIndex)
        }

        return iterator
    }

    static func firstRawValue() -> Int {
        return 0
    }
}

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)
}

在处理Swift 2.0时,以下是我的建议:

我已经将原始类型添加到Suit enum

enum Suit: Int {

然后:

struct Card {
    var rank: Rank
    var suit: Suit


    func fullDeck()-> [Card] {

        var deck = [Card]()

        for i in Rank.Ace.rawValue...Rank.King.rawValue {

            for j in Suit.Spades.rawValue...Suit.Clubs.rawValue {

                deck.append(Card(rank:Rank(rawValue: i)! , suit: Suit(rawValue: j)!))
            }
        }

        return deck
    }
}