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"]
其他回答
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
}
}
有时,您可能会处理具有底层原始整数类型的枚举类型,这种类型在整个软件开发生命周期中都会发生变化。下面是一个很适合这种情况的例子:
public class MyClassThatLoadsTexturesEtc
{
//...
// Colors used for gems and sectors.
public enum Color: Int
{
// Colors arranged in order of the spectrum.
case First = 0
case Red, Orange, Yellow, Green, Blue, Purple, Pink
// --> Add more colors here, between the first and last markers.
case Last
}
//...
public func preloadGems()
{
// Preload all gems.
for i in (Color.First.toRaw() + 1) ..< (Color.Last.toRaw())
{
let color = Color.fromRaw(i)!
loadColoredTextures(forKey: color)
}
}
//...
}
在Swift中,枚举类型可以像EnumType一样访问。案例:
let tableView = UITableView(frame: self.view. view)UITableViewStyle.Plain)
大多数情况下,只有当您有几个选项可以使用,并且确切地知道在每个选项上要做什么时,才会使用枚举类型。
在处理枚举类型时,使用for-in结构没有太大意义。
你可以这样做,例如:
func sumNumbers(numbers : Int...) -> Int {
var sum = 0
for number in numbers{
sum += number
}
return sum
}
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
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