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
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
}
}
其他回答
有一种聪明的方法，尽管令人沮丧，但它说明了两种不同类型的枚举之间的区别。
试试这个:
func makeDeck() -> Card[] {
var deck: Card[] = []
var suits: Suit[] = [.Hearts, .Diamonds, .Clubs, .Spades]
for i in 1...13 {
for suit in suits {
deck += Card(rank: Rank.fromRaw(i)!, suit: suit)
}
}
return deck
}
交易是，一个由数字(原始值)支持的枚举是隐式显式有序的，而一个没有数字支持的枚举是显式隐式无序的。
例如，当我们给枚举值数字时，语言足够狡猾，可以找出数字的顺序。 另一方面，如果我们不给它任何顺序，当我们尝试迭代这些值时，语言就会举起双手说:“是的，但你想先执行哪个??”
其他可以做到这一点(迭代无序枚举)的语言可能是相同的语言，其中所有内容实际上都是一个地图或字典，你可以迭代地图的键，无论是否有任何逻辑顺序。
诀窍是给它提供一些显式排序的东西，在这个例子中，suit的实例在数组中按照我们想要的顺序。一旦你这么说，霉霉就会说“你为什么不一开始就这么说呢?”
另一个简写技巧是在fromRaw函数上使用强制操作符。这说明了关于枚举的另一个“陷阱”，即可能传入的值的范围通常大于枚举的范围。例如，如果我们说Rank.fromRaw(60)，就不会返回值，所以我们使用了语言的可选特性，在我们开始使用可选特性的地方，很快就会出现强制。(或者交替if let结构，这对我来说仍然有点奇怪)
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
}
}
实验内容是: 实验
在Card中添加一个方法，用于创建一副完整的牌，每一副牌都是rank和花色的组合。
因此，除了添加方法之外，没有修改或增强给定的代码(并且没有使用还没有教过的东西)，我想出了这个解决方案:
struct Card {
var rank: Rank
var suit: Suit
func simpleDescription() -> String {
return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
}
func createDeck() -> [Card] {
var deck: [Card] = []
for rank in Rank.Ace.rawValue...Rank.King.rawValue {
for suit in Suit.Spades.rawValue...Suit.Clubs.rawValue {
let card = Card(rank: Rank(rawValue: rank)!, suit: Suit(rawValue: suit)!)
//println(card.simpleDescription())
deck += [card]
}
}
return deck
}
}
let threeOfSpades = Card(rank: .Three, suit: .Spades)
let threeOfSpadesDescription = threeOfSpades.simpleDescription()
let deck = threeOfSpades.createDeck()
这篇文章是相关的https://www.swift-studies.com/blog/2014/6/10/enumerating-enums-in-swift
基本上，提议的解决方案是
enum ProductCategory : String {
case Washers = "washers", Dryers = "dryers", Toasters = "toasters"
static let allValues = [Washers, Dryers, Toasters]
}
for category in ProductCategory.allValues{
//Do something
}
这看起来像一个黑客，但如果你使用原始值，你可以这样做
enum Suit: Int {
case Spades = 0, Hearts, Diamonds, Clubs
...
}
var suitIndex = 0
while var suit = Suit.fromRaw(suitIndex++) {
...
}
