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
您可以通过实现ForwardIndexType协议来迭代枚举。
ForwardIndexType协议要求您定义一个继任者()函数来逐级遍历元素。
enum Rank: Int, ForwardIndexType {
case Ace = 1
case Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten
case Jack, Queen, King
// ... other functions
// Option 1 - Figure it out by hand
func successor() -> Rank {
switch self {
case .Ace:
return .Two
case .Two:
return .Three
// ... etc.
default:
return .King
}
}
// Option 2 - Define an operator!
func successor() -> Rank {
return self + 1
}
}
// NOTE: The operator is defined OUTSIDE the class
func + (left: Rank, right: Int) -> Rank {
// I'm using to/from raw here, but again, you can use a case statement
// or whatever else you can think of
return left == .King ? .King : Rank(rawValue: left.rawValue + right)!
}
在开或闭范围内迭代(..<或…)将在内部调用继任者()函数，允许你这样写:
// Under the covers, successor(Rank.King) and successor(Rank.Ace) are called to establish limits
for r in Rank.Ace...Rank.King {
// Do something useful
}
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"]
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++)
}
}
)
}
}
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?)
我在比特和字节中跌跌撞撞，创建了一个扩展，后来我发现它与@rintaro的答案非常相似。它是这样使用的:
enum E : EnumCollection {
case A, B, C
}
Array(E.cases()) // [A, B, C]
值得注意的是，它可以在任何没有关联值的enum上使用。注意，这对于没有大小写的枚举不起作用。
与@rintaro的答案一样，这段代码使用枚举的底层表示。这种表示没有文档化，将来可能会改变，这会破坏它。我不建议在生产中使用这种方法。
代码(Swift 2.2, Xcode 7.3.1，不工作在Xcode 10):
protocol EnumCollection : Hashable {}
extension EnumCollection {
static func cases() -> AnySequence<Self> {
typealias S = Self
return 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
}
}
}
}
代码(Swift 3, Xcode 8.1，不工作在Xcode 10):
protocol EnumCollection : Hashable {}
extension EnumCollection {
static func cases() -> AnySequence<Self> {
typealias S = Self
return AnySequence { () -> AnyIterator<S> in
var raw = 0
return AnyIterator {
let current : Self = withUnsafePointer(to: &raw) { $0.withMemoryRebound(to: S.self, capacity: 1) { $0.pointee } }
guard current.hashValue == raw else { return nil }
raw += 1
return current
}
}
}
}
我不知道为什么我需要typealias，但编译器抱怨没有它。
