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:
♠
♥
♦
♣
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
其他回答
这个问题现在简单多了。以下是我的Swift 4.2解决方案:
enum Suit: Int, CaseIterable {
case None
case Spade, Heart, Diamond, Club
static let allNonNullCases = Suit.allCases[Spade.rawValue...]
}
enum Rank: Int, CaseIterable {
case Joker
case Two, Three, Four, Five, Six, Seven, Eight
case Nine, Ten, Jack, Queen, King, Ace
static let allNonNullCases = Rank.allCases[Two.rawValue...]
}
func makeDeck(withJoker: Bool = false) -> [Card] {
var deck = [Card]()
for suit in Suit.allNonNullCases {
for rank in Rank.allNonNullCases {
deck.append(Card(suit: suit, rank: rank))
}
}
if withJoker {
deck.append(Card(suit: .None, rank: .Joker))
}
return deck
}
4。2:
我喜欢这个解决方案,我把找到“列表理解在Swift”。
它使用Int rawws而不是string,但它避免了键入两次,它允许自定义范围,并且不硬编码原始值。
这是我最初解决方案的Swift 4版本,但请参阅上面的4.2改进:
enum Suit: Int {
case None
case Spade, Heart, Diamond, Club
static let allRawValues = Suit.Spade.rawValue...Suit.Club.rawValue
static let allCases = Array(allRawValues.map{ Suit(rawValue: $0)! })
}
enum Rank: Int {
case Joker
case Two, Three, Four, Five, Six
case Seven, Eight, Nine, Ten
case Jack, Queen, King, Ace
static let allRawValues = Rank.Two.rawValue...Rank.Ace.rawValue
static let allCases = Array(allRawValues.map{ Rank(rawValue: $0)! })
}
func makeDeck(withJoker: Bool = false) -> [Card] {
var deck = [Card]()
for suit in Suit.allCases {
for rank in Rank.allCases {
deck.append(Card(suit: suit, rank: rank))
}
}
if withJoker {
deck.append(Card(suit: .None, rank: .Joker))
}
return deck
}
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
}
}
Swift 4 + 2。
从Swift 4.2 (Xcode 10)开始,只需将协议一致性添加到CaseIterable中,就可以从allCases中受益。要添加这个协议一致性,你只需要在某个地方写:
extension Suit: CaseIterable {}
如果枚举是你自己的,你可以直接在声明中指定一致性:
enum Suit: String, CaseIterable { case spades = "♠"; case hearts = "♥"; case diamonds = "♦"; case clubs = "♣" }
然后下面的代码将打印所有可能的值:
Suit.allCases.forEach {
print($0.rawValue)
}
与早期Swift版本的兼容性(3。X和4.x)
如果您需要支持Swift 3。x或4.0,你可以通过添加以下代码来模仿Swift 4.2的实现:
#if !swift(>=4.2)
public protocol CaseIterable {
associatedtype AllCases: Collection where AllCases.Element == Self
static var allCases: AllCases { get }
}
extension CaseIterable where Self: Hashable {
static var allCases: [Self] {
return [Self](AnySequence { () -> AnyIterator<Self> in
var raw = 0
var first: Self?
return AnyIterator {
let current = withUnsafeBytes(of: &raw) { $0.load(as: Self.self) }
if raw == 0 {
first = current
} else if current == first {
return nil
}
raw += 1
return current
}
})
}
}
#endif
以下是我的建议。这不是完全令人满意的(我对Swift和OOP很陌生!),但也许有人可以改进它。这个想法是让每个枚举提供自己的范围信息作为.first和.last属性。它只向每个枚举添加了两行代码:仍然有点硬编码,但至少它没有复制整个集合。它确实需要将Suit enum修改为Int类型,就像Rank enum一样,而不是无类型的。
而不是重复整个解决方案,下面是我添加到。在case语句之后的某个地方(Suit enum类似):
var first: Int { return Ace.toRaw() }
var last: Int { return King.toRaw() }
以及我用来将deck构建为String数组的循环。(问题定义没有说明牌组是如何构造的。)
func createDeck() -> [String] {
var deck: [String] = []
var card: String
for r in Rank.Ace.first...Rank.Ace.last {
for s in Suit.Hearts.first...Suit.Hearts.last {
card = Rank.simpleDescription( Rank.fromRaw(r)!)() + " of " + Suit.simpleDescription( Suit.fromRaw(s)!)()
deck.append( card)
}
}
return deck
}
这并不令人满意,因为属性与元素而不是enum相关联。但它确实为“for”循环增加了清晰度。我希望它是Rank。而不是Rank.Ace.first。它适用于任何元素,但很难看。有人能演示一下如何将其提升到enum级别吗?
为了使它工作,我从Card结构中提取了createDeck方法。我不知道如何从该结构返回一个[String]数组,这似乎是一个糟糕的地方,把这样的方法无论如何。
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”]