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
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++)
}
}
)
}
}
其他回答
它花了我一点，而不仅仅是一个方法在结构像swift书调用，但我在枚举中设置了下一个函数。我会使用一个协议，我不知道为什么，但有秩设置为int混乱。
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.toRaw())
}
}
mutating func next() -> Rank {
var rank = self
var rawrank = rank.toRaw()
var nrank: Rank = self
rawrank = rawrank + 1
if let newRank = Rank.fromRaw(rawrank) {
println("\(newRank.simpleDescription())")
nrank = newRank
} else {
return self
}
return nrank
}
}
enum Suit {
case Spades, Hearts, Diamonds, Clubs
func color() -> String {
switch self {
case .Spades, .Clubs:
return "black"
default:
return "red"
}
}
func simpleDescription() -> String {
switch self {
case .Spades:
return "spades"
case .Hearts:
return "hearts"
case .Diamonds:
return "diamonds"
case .Clubs:
return "clubs"
}
}
mutating func next() -> Suit {
switch self {
case .Spades:
return Hearts
case .Hearts:
return Diamonds
case .Diamonds:
return Clubs
case .Clubs:
return Spades
}
}
}
struct Card {
var rank: Rank
var suit: Suit
func deck() -> Card[] {
var tRank = self.rank
var tSuit = self.suit
let tcards = 52 // we start from 0
var cards: Card[] = []
for i in 0..tcards {
var card = Card(rank: tRank, suit: tSuit)
cards.append(card)
tRank = tRank.next()
tSuit = tSuit.next()
}
return cards
}
func simpleDescription() -> String {
return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
}
}
var card = Card(rank: .Ace, suit: .Spades)
var deck = card.deck()
我使用了一些常识，但这可以通过将花色乘以等级来轻松纠正(如果你没有使用标准的桥牌，你必须相应地改变枚举，如果基本上只是通过不同的枚举进行步骤)。为了节省时间，我使用了ranks rawValues，如果你愿意，你也可以为西装做同样的事情。然而，这个例子没有它，所以我决定在不改变suit rawValue的情况下找出它
Xcode 10与Swift 4.2
enum Filter: String, CaseIterable {
case salary = "Salary"
case experience = "Experience"
case technology = "Technology"
case unutilized = "Unutilized"
case unutilizedHV = "Unutilized High Value"
static let allValues = Filter.allCases.map { $0.rawValue }
}
叫它
print(Filter.allValues)
打印:
[“薪酬”、“经验”、“技术”、“未利用”、“未利用的高价值”]
旧版本
对于表示Int的enum
enum Filter: Int {
case salary
case experience
case technology
case unutilized
case unutilizedHV
static let allRawValues = salary.rawValue...unutilizedHV.rawValue // First to last case
static let allValues = allRawValues.map { Filter(rawValue: $0)!.rawValue }
}
这样叫它:
print(Filter.allValues)
打印:
[0, 1, 2, 3, 4]
用于表示字符串的enum
enum Filter: Int {
case salary
case experience
case technology
case unutilized
case unutilizedHV
static let allRawValues = salary.rawValue...unutilizedHV.rawValue // First to last case
static let allValues = allRawValues.map { Filter(rawValue: $0)!.description }
}
extension Filter: CustomStringConvertible {
var description: String {
switch self {
case .salary: return "Salary"
case .experience: return "Experience"
case .technology: return "Technology"
case .unutilized: return "Unutilized"
case .unutilizedHV: return "Unutilized High Value"
}
}
}
叫它
print(Filter.allValues)
打印:
[“薪酬”、“经验”、“技术”、“未利用”、“未利用的高价值”]
原则上，如果你不为enum的大小写使用原始值赋值，这样做是可能的:
enum RankEnum: Int {
case Ace
case One
case Two
}
class RankEnumGenerator: Generator {
var i = 0
typealias Element = RankEnum
func next() -> Element? {
let r = RankEnum.fromRaw(i)
i += 1
return r
}
}
extension RankEnum {
static func enumerate() -> SequenceOf<RankEnum> {
return SequenceOf<RankEnum>({ RankEnumGenerator() })
}
}
for r in RankEnum.enumerate() {
println("\(r.toRaw())")
}
我创建了一个实用函数iterateEnum()，用于迭代任意枚举类型的情况。
下面是示例用法:
enum Suit: String {
case Spades = "♠"
case Hearts = "♥"
case Diamonds = "♦"
case Clubs = "♣"
}
for f in iterateEnum(Suit) {
println(f.rawValue)
}
输出:
♠
♥
♦
♣
但是，这仅用于调试或测试目的:这依赖于几个未记录的Swift1.1编译器行为，因此，使用它的风险由您自己承担。
代码如下:
func iterateEnum<T: Hashable>(_: T.Type) -> GeneratorOf<T> {
var cast: (Int -> T)!
switch sizeof(T) {
case 0: return GeneratorOf(GeneratorOfOne(unsafeBitCast((), T.self)))
case 1: cast = { unsafeBitCast(UInt8(truncatingBitPattern: $0), T.self) }
case 2: cast = { unsafeBitCast(UInt16(truncatingBitPattern: $0), T.self) }
case 4: cast = { unsafeBitCast(UInt32(truncatingBitPattern: $0), T.self) }
case 8: cast = { unsafeBitCast(UInt64($0), T.self) }
default: fatalError("cannot be here")
}
var i = 0
return GeneratorOf {
let next = cast(i)
return next.hashValue == i++ ? next : nil
}
}
其基本思想是:
枚举的内存表示，不包括有关联类型的枚举，只是一个案例的索引，当案例的计数是2…256，它和UInt8是一样的，当257…65536，它是UInt16等等。因此，它可以是unsafeBitcast对应的无符号整数类型。 枚举值的. hashvalue与case的索引相同。 从无效索引位转换的枚举值的. hashvalue为0。
为Swift2修改，并从@Kametrixom的回答中实现了选角想法:
func iterateEnum<T: Hashable>(_: T.Type) -> AnyGenerator<T> {
var i = 0
return anyGenerator {
let next = withUnsafePointer(&i) { UnsafePointer<T>($0).memory }
return next.hashValue == i++ ? next : nil
}
}
对Swift3的修订:
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
}
}
针对Swift3.0.1修订:
func iterateEnum<T: Hashable>(_: T.Type) -> AnyIterator<T> {
var i = 0
return AnyIterator {
let next = withUnsafeBytes(of: &i) { $0.load(as: T.self) }
if next.hashValue != i { return nil }
i += 1
return next
}
}
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()
