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
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
}
}
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"]
你可以试着像这样列举
enum Planet: String {
case Mercury
case Venus
case Earth
case Mars
static var enumerate: [Planet] {
var a: [Planet] = []
switch Planet.Mercury {
case .Mercury: a.append(.Mercury); fallthrough
case .Venus: a.append(.Venus); fallthrough
case .Earth: a.append(.Earth); fallthrough
case .Mars: a.append(.Mars)
}
return a
}
}
Planet.enumerate // [Mercury, Venus, Earth, Mars]
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)")
}
enum Rank: Int {
...
static let ranks = (Rank.Ace.rawValue ... Rank.King.rawValue).map{Rank(rawValue: $0)! }
}
enum Suit {
...
static let suits = [Spades, Hearts, Diamonds, Clubs]
}
struct Card {
...
static func fullDesk() -> [Card] {
var desk: [Card] = []
for suit in Suit.suits {
for rank in Rank.ranks {
desk.append(Card(rank: rank,suit: suit))
}
}
return desk
}
}
这个怎么样?