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
我创建了一个实用函数iterateEnum(),用于迭代任意枚举类型的情况。
下面是示例用法:
enum Suit: String {
case Spades = "♠"
case Hearts = "♥"
case Diamonds = "♦"
case Clubs = "♣"
}
for f in iterateEnum(Suit) {
println(f.rawValue)
}
输出:
♠
♥
♦
♣
但是,这仅用于调试或测试目的:这依赖于几个未记录的Swift1.1编译器行为,因此,使用它的风险由您自己承担。
代码如下:
func iterateEnum<T: Hashable>(_: T.Type) -> GeneratorOf<T> {
var cast: (Int -> T)!
switch sizeof(T) {
case 0: return GeneratorOf(GeneratorOfOne(unsafeBitCast((), T.self)))
case 1: cast = { unsafeBitCast(UInt8(truncatingBitPattern: $0), T.self) }
case 2: cast = { unsafeBitCast(UInt16(truncatingBitPattern: $0), T.self) }
case 4: cast = { unsafeBitCast(UInt32(truncatingBitPattern: $0), T.self) }
case 8: cast = { unsafeBitCast(UInt64($0), T.self) }
default: fatalError("cannot be here")
}
var i = 0
return GeneratorOf {
let next = cast(i)
return next.hashValue == i++ ? next : nil
}
}
其基本思想是:
枚举的内存表示,不包括有关联类型的枚举,只是一个案例的索引,当案例的计数是2…256,它和UInt8是一样的,当257…65536,它是UInt16等等。因此,它可以是unsafeBitcast对应的无符号整数类型。 枚举值的. hashvalue与case的索引相同。 从无效索引位转换的枚举值的. hashvalue为0。
为Swift2修改,并从@Kametrixom的回答中实现了选角想法:
func iterateEnum<T: Hashable>(_: T.Type) -> AnyGenerator<T> {
var i = 0
return anyGenerator {
let next = withUnsafePointer(&i) { UnsafePointer<T>($0).memory }
return next.hashValue == i++ ? next : nil
}
}
对Swift3的修订:
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
}
}
针对Swift3.0.1修订:
func iterateEnum<T: Hashable>(_: T.Type) -> AnyIterator<T> {
var i = 0
return AnyIterator {
let next = withUnsafeBytes(of: &i) { $0.load(as: T.self) }
if next.hashValue != i { return nil }
i += 1
return next
}
}
其他回答
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)
}
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()
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
}
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
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)")
}