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
编辑: 快速进化提案SE-0194枚举案例派生集合为这个问题提出了一个水平的解决方案。我们在Swift 4.2和更新版本中看到了它。该提案还指出了一些变通方法，这些方法与这里已经提到的一些方法类似，但可能会很有趣。
为了完整起见，我也会保留我原来的职位。
这是基于@Peymmankh的回答的另一种方法，适用于Swift 3。
public protocol EnumCollection: Hashable {}
extension EnumCollection {
public static func allValues() -> [Self] {
typealias S = Self
let retVal = AnySequence { () -> AnyIterator<S> in
var raw = 0
return AnyIterator {
let current = withUnsafePointer(to: &raw) {
$0.withMemoryRebound(to: S.self, capacity: 1) { $0.pointee }
}
guard current.hashValue == raw else { return nil }
raw += 1
return current
}
}
return [S](retVal)
}
其他回答
与@Kametrixom的答案在这里，我相信返回一个数组将比返回AnySequence更好，因为你可以访问所有数组的好东西，如计数等。
以下是改写后的内容:
public protocol EnumCollection : Hashable {}
extension EnumCollection {
public static func allValues() -> [Self] {
typealias S = Self
let retVal = 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
}
}
return [S](retVal)
}
}
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()
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
}
下面是我用来迭代枚举和从一个枚举提供多个值类型的方法
enum IterateEnum: Int {
case Zero
case One
case Two
case Three
case Four
case Five
case Six
case Seven
//tuple allows multiple values to be derived from the enum case, and
//since it is using a switch with no default, if a new case is added,
//a compiler error will be returned if it doesn't have a value tuple set
var value: (french: String, spanish: String, japanese: String) {
switch self {
case .Zero: return (french: "zéro", spanish: "cero", japanese: "nuru")
case .One: return (french: "un", spanish: "uno", japanese: "ichi")
case .Two: return (french: "deux", spanish: "dos", japanese: "ni")
case .Three: return (french: "trois", spanish: "tres", japanese: "san")
case .Four: return (french: "quatre", spanish: "cuatro", japanese: "shi")
case .Five: return (french: "cinq", spanish: "cinco", japanese: "go")
case .Six: return (french: "six", spanish: "seis", japanese: "roku")
case .Seven: return (french: "sept", spanish: "siete", japanese: "shichi")
}
}
//Used to iterate enum or otherwise access enum case by index order.
//Iterate by looping until it returns nil
static func item(index: Int) -> IterateEnum? {
return IterateEnum.init(rawValue: index)
}
static func numberFromSpanish(number: String) -> IterateEnum? {
return findItem { $0.value.spanish == number }
}
//use block to test value property to retrieve the enum case
static func findItem(predicate: ((_: IterateEnum) -> Bool)) -> IterateEnum? {
var enumIndex: Int = -1
var enumCase: IterateEnum?
//Iterate until item returns nil
repeat {
enumIndex += 1
enumCase = IterateEnum.item(index: enumIndex)
if let eCase = enumCase {
if predicate(eCase) {
return eCase
}
}
} while enumCase != nil
return nil
}
}
var enumIndex: Int = -1
var enumCase: IterateEnum?
// Iterate until item returns nil
repeat {
enumIndex += 1
enumCase = IterateEnum.item(index: enumIndex)
if let eCase = enumCase {
print("The number \(eCase) in french: \(eCase.value.french), spanish: \(eCase.value.spanish), japanese: \(eCase.value.japanese)")
}
} while enumCase != nil
print("Total of \(enumIndex) cases")
let number = IterateEnum.numberFromSpanish(number: "siete")
print("siete in japanese: \((number?.value.japanese ?? "Unknown"))")
输出如下:
法语中的数字Zero: zéro，西班牙语中的数字cero，日语中的数字nuru 数字一在法语中是un，西班牙语中是uno，日语中是ichi 法语中的数字2是deux，西班牙语中的数字2是dos，日语中的数字2是ni 法语中的“三”是“trois”，西班牙语中的“tres”，日语中的“san” 法语中的“四”是quatre，西班牙语中的“四”是cuatro，日语中的“四”是shi 数字五在法语中是cinq，西班牙语中是cinco，日语中是go 数字6在法语中是Six，西班牙语是seis，日语是roku 法语中的数字“七”是“sept”，西班牙语中的“siete”，日语中的“shichi”
共8例
Siete在日语中的意思是:shichi
更新
我最近创建了一个协议来处理枚举。该协议需要一个Int原始值的enum:
protocol EnumIteration {
//Used to iterate enum or otherwise access enum case by index order. Iterate by looping until it returns nil
static func item(index:Int) -> Self?
static func iterate(item:((index:Int, enumCase:Self)->()), completion:(()->())?) {
static func findItem(predicate:((enumCase:Self)->Bool)) -> Self?
static func count() -> Int
}
extension EnumIteration where Self: RawRepresentable, Self.RawValue == Int {
//Used to iterate enum or otherwise access enum case by index order. Iterate by looping until it returns nil
static func item(index:Int) -> Self? {
return Self.init(rawValue: index)
}
static func iterate(item:((index:Int, enumCase:Self)->()), completion:(()->())?) {
var enumIndex:Int = -1
var enumCase:Self?
//Iterate until item returns nil
repeat {
enumIndex += 1
enumCase = Self.item(enumIndex)
if let eCase = enumCase {
item(index: enumIndex, enumCase: eCase)
}
} while enumCase != nil
completion?()
}
static func findItem(predicate:((enumCase:Self)->Bool)) -> Self? {
var enumIndex:Int = -1
var enumCase:Self?
//Iterate until item returns nil
repeat {
enumIndex += 1
enumCase = Self.item(enumIndex)
if let eCase = enumCase {
if predicate(enumCase:eCase) {
return eCase
}
}
} while enumCase != nil
return nil
}
static func count() -> Int {
var enumIndex:Int = -1
var enumCase:Self?
//Iterate until item returns nil
repeat {
enumIndex += 1
enumCase = Self.item(enumIndex)
} while enumCase != nil
//last enumIndex (when enumCase == nil) is equal to the enum count
return enumIndex
}
}
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())
}
}
}
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, .Clubs: return "black"
case .Hearts, .Diamonds: return "red"
}
}
}
struct Card {
var rank: Rank
var suit: Suit
func simpleDescription() -> String {
return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
}
static func createPokers() -> Card[] {
let ranks = Array(Rank.Ace.toRaw()...Rank.King.toRaw())
let suits = Array(Suit.Spades.toRaw()...Suit.Clubs.toRaw())
let cards = suits.reduce(Card[]()) { (tempCards, suit) in
tempCards + ranks.map { rank in
Card(rank: Rank.fromRaw(rank)!, suit: Suit.fromRaw(suit)!)
}
}
return cards
}
}
