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
Swift 4 + 2。
从Swift 4.2 (Xcode 10)开始,只需将协议一致性添加到CaseIterable中,就可以从allCases中受益。要添加这个协议一致性,你只需要在某个地方写:
extension Suit: CaseIterable {}
如果枚举是你自己的,你可以直接在声明中指定一致性:
enum Suit: String, CaseIterable { case spades = "♠"; case hearts = "♥"; case diamonds = "♦"; case clubs = "♣" }
然后下面的代码将打印所有可能的值:
Suit.allCases.forEach {
print($0.rawValue)
}
与早期Swift版本的兼容性(3。X和4.x)
如果您需要支持Swift 3。x或4.0,你可以通过添加以下代码来模仿Swift 4.2的实现:
#if !swift(>=4.2)
public protocol CaseIterable {
associatedtype AllCases: Collection where AllCases.Element == Self
static var allCases: AllCases { get }
}
extension CaseIterable where Self: Hashable {
static var allCases: [Self] {
return [Self](AnySequence { () -> AnyIterator<Self> in
var raw = 0
var first: Self?
return AnyIterator {
let current = withUnsafeBytes(of: &raw) { $0.load(as: Self.self) }
if raw == 0 {
first = current
} else if current == first {
return nil
}
raw += 1
return current
}
})
}
}
#endif
其他回答
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
}
在Swift 3中,当底层枚举有rawValue时,你可以实现Strideable协议。优点是不像其他建议那样创建值数组,并且标准的Swift“for in”循环工作,这是一个很好的语法。
// "Int" to get rawValue, and Strideable so we can iterate
enum MyColorEnum: Int, Strideable {
case Red
case Green
case Blue
case Black
// required by Strideable
typealias Stride = Int
func advanced(by n:Stride) -> MyColorEnum {
var next = self.rawValue + n
if next > MyColorEnum.Black.rawValue {
next = MyColorEnum.Black.rawValue
}
return MyColorEnum(rawValue: next)!
}
func distance(to other: MyColorEnum) -> Int {
return other.rawValue - self.rawValue
}
// just for printing
func simpleDescription() -> String {
switch self {
case .Red: return "Red"
case .Green: return "Green"
case .Blue: return "Blue"
case .Black: return "Black"
}
}
}
// this is how you use it:
for i in MyColorEnum.Red ... MyColorEnum.Black {
print("ENUM: \(i)")
}
Swift 5解决方案:
enum Suit: String, CaseIterable {
case spades = "♠"
case hearts = "♥"
case diamonds = "♦"
case clubs = "♣"
}
// access cases like this:
for suitKey in Suit.allCases {
print(suitKey)
}
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"]
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
}