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
我使用了下面的方法,假设我知道哪个是Rank enum中的最后一个值,所有的Rank在Ace之后都有增量值
我喜欢这种方式,因为它干净,小,容易理解
func cardDeck() -> Card[] {
var cards: Card[] = []
let minRank = Rank.Ace.toRaw()
let maxRank = Rank.King.toRaw()
for rank in minRank...maxRank {
if var convertedRank: Rank = Rank.fromRaw(rank) {
cards.append(Card(rank: convertedRank, suite: Suite.Clubs))
cards.append(Card(rank: convertedRank, suite: Suite.Diamonds))
cards.append(Card(rank: convertedRank, suite: Suite.Hearts))
cards.append(Card(rank: convertedRank, suite: Suite.Spades))
}
}
return cards
}
其他回答
我的解决方案是声明一个包含所有枚举可能性的数组。所以for循环可以遍历所有这些。
//Function inside struct Card
static func generateFullDeck() -> [Card] {
let allRanks = [Rank.Ace, Rank.Two, Rank.Three, Rank.Four, Rank.Five, Rank.Six, Rank.Seven, Rank.Eight, Rank.Nine, Rank.Ten, Rank.Jack, Rank.Queen, Rank.King]
let allSuits = [Suit.Hearts, Suit.Diamonds, Suit.Clubs, Suit.Spades]
var myFullDeck: [Card] = []
for myRank in allRanks {
for mySuit in allSuits {
myFullDeck.append(Card(rank: myRank, suit: mySuit))
}
}
return myFullDeck
}
//actual use:
let aFullDeck = Card.generateFullDeck() //Generate the desired full deck
var allDesc: [String] = []
for aCard in aFullDeck {
println(aCard.simpleDescription()) //You'll see all the results in playground
}
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"]
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
}
}
(改进Karthik Kumar的回答)
这个解决方案是使用编译器来保证你不会错过一个case。
enum Suit: String {
case spades = "♠"
case hearts = "♥"
case diamonds = "♦"
case clubs = "♣"
static var enumerate: [Suit] {
switch Suit.spades {
// make sure the two lines are identical ^_^
case .spades, .hearts, .diamonds, .clubs:
return [.spades, .hearts, .diamonds, .clubs]
}
}
}
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)")
}