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
另一个解决方案:
enum Suit: String {
case spades = "♠"
case hearts = "♥"
case diamonds = "♦"
case clubs = "♣"
static var count: Int {
return 4
}
init(index: Int) {
switch index {
case 0: self = .spades
case 1: self = .hearts
case 2: self = .diamonds
default: self = .clubs
}
}
}
for i in 0..<Suit.count {
print(Suit(index: i).rawValue)
}
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
}
}
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
}
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
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
}
}
在处理Swift 2.0时，以下是我的建议:
我已经将原始类型添加到Suit enum
enum Suit: Int {
然后:
struct Card {
var rank: Rank
var suit: Suit
func fullDeck()-> [Card] {
var deck = [Card]()
for i in Rank.Ace.rawValue...Rank.King.rawValue {
for j in Suit.Spades.rawValue...Suit.Clubs.rawValue {
deck.append(Card(rank:Rank(rawValue: i)! , suit: Suit(rawValue: j)!))
}
}
return deck
}
}
