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
我发现了一种有点俗气但更安全的方法,它不需要键入两次值或引用枚举值的内存,因此不太可能损坏。
基本上,与其使用枚举,不如创建一个具有单个实例的结构体,并将所有enum-values设置为常量。然后可以使用Mirror查询变量
public struct Suit{
// the values
let spades = "♠"
let hearts = "♥"
let diamonds = "♦"
let clubs = "♣"
// make a single instance of the Suit struct, Suit.instance
struct SStruct{static var instance: Suit = Suit()}
static var instance : Suit{
get{return SStruct.instance}
set{SStruct.instance = newValue}
}
// an array with all of the raw values
static var allValues: [String]{
var values = [String]()
let mirror = Mirror(reflecting: Suit.instance)
for (_, v) in mirror.children{
guard let suit = v as? String else{continue}
values.append(suit)
}
return values
}
}
如果使用此方法,则需要使用Suit.instance.clubs或Suit.instance.spades来获取单个值
但所有这些都太无聊了……让我们做一些事情,使它更像一个真正的enum!
public struct SuitType{
// store multiple things for each suit
let spades = Suit("♠", order: 4)
let hearts = Suit("♥", order: 3)
let diamonds = Suit("♦", order: 2)
let clubs = Suit("♣", order: 1)
struct SStruct{static var instance: SuitType = SuitType()}
static var instance : SuitType{
get{return SStruct.instance}
set{SStruct.instance = newValue}
}
// a dictionary mapping the raw values to the values
static var allValuesDictionary: [String : Suit]{
var values = [String : Suit]()
let mirror = Mirror(reflecting: SuitType.instance)
for (_, v) in mirror.children{
guard let suit = v as? Suit else{continue}
values[suit.rawValue] = suit
}
return values
}
}
public struct Suit: RawRepresentable, Hashable{
public var rawValue: String
public typealias RawValue = String
public var hashValue: Int{
// find some integer that can be used to uniquely identify
// each value. In this case, we could have used the order
// variable because it is a unique value, yet to make this
// apply to more cases, the hash table address of rawValue
// will be returned, which should work in almost all cases
//
// you could also add a hashValue parameter to init() and
// give each suit a different hash value
return rawValue.hash
}
public var order: Int
public init(_ value: String, order: Int){
self.rawValue = value
self.order = order
}
// an array of all of the Suit values
static var allValues: [Suit]{
var values = [Suit]()
let mirror = Mirror(reflecting: SuitType.instance)
for (_, v) in mirror.children{
guard let suit = v as? Suit else{continue}
values.append(suit)
}
return values
}
// allows for using Suit(rawValue: "♦"), like a normal enum
public init?(rawValue: String){
// get the Suit from allValuesDictionary in SuitType, or return nil if that raw value doesn't exist
guard let suit = SuitType.allValuesDictionary[rawValue] else{return nil}
// initialize a new Suit with the same properties as that with the same raw value
self.init(suit.rawValue, order: suit.order)
}
}
你现在可以做
let allSuits: [Suit] = Suit.allValues
or
for suit in Suit.allValues{
print("The suit \(suit.rawValue) has the order \(suit.order)")
}
然而,要获得一个单一,你仍然需要使用SuitType.instance.spades或SuitType.instance.hearts。为了更加直观,您可以向Suit添加一些允许您使用Suit.type的代码。*而不是SuitType.instance.*
public struct Suit: RawRepresentable, Hashable{
// ...your code...
static var type = SuitType.instance
// ...more of your code...
}
您现在可以使用Suit.type.diamonds而不是SuitType.instance。diamonds,或者Suit.type.clubs而不是SuitType.instance.clubs
其他回答
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)
}
}
//...
}
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()
更新到Swift 2.2+
func iterateEnum<T: Hashable>(_: T.Type) -> AnyGenerator<T> {
var i = 0
return AnyGenerator {
let next = withUnsafePointer(&i) {
UnsafePointer<T>($0).memory
}
if next.hashValue == i {
i += 1
return next
} else {
return nil
}
}
}
它更新了Swift 2.2表单@Kametrixom的答案
Swift 3.0+(非常感谢@Philip)
func iterateEnum<T: Hashable>(_: T.Type) -> AnyIterator<T> {
var i = 0
return AnyIterator {
let next = withUnsafePointer(&i) {
UnsafePointer<T>($0).pointee
}
if next.hashValue == i {
i += 1
return next
} else {
return nil
}
}
}
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++) {
...
}
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
}