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
enum Rank: Int
{
case Ace = 0
case Two, Three, Four, Five, Six, Seve, Eight, Nine, Ten
case Jack, Queen, King
case Count
}
enum Suit : Int
{
case Spades = 0
case Hearts, Diamonds, Clubs
case Count
}
struct Card
{
var rank:Rank
var suit:Suit
}
class Test
{
func makeDeck() -> Card[]
{
let suitsCount:Int = Suit.Count.toRaw()
let rankCount:Int = Rank.Count.toRaw()
let repeatedCard:Card = Card(rank:Rank.Ace, suit:Suit.Spades)
let deck:Card[] = Card[](count:suitsCount*rankCount, repeatedValue:repeatedCard)
for i:Int in 0..rankCount
{
for j:Int in 0..suitsCount
{
deck[i*suitsCount+j] = Card(rank: Rank.fromRaw(i)!, suit: Suit.fromRaw(j)!)
}
}
return deck
}
}
根据Rick的回答:这要快5倍
其他回答
编辑: 快速进化提案SE-0194枚举案例派生集合为这个问题提出了一个水平的解决方案。我们在Swift 4.2和更新版本中看到了它。该提案还指出了一些变通方法,这些方法与这里已经提到的一些方法类似,但可能会很有趣。
为了完整起见,我也会保留我原来的职位。
这是基于@Peymmankh的回答的另一种方法,适用于Swift 3。
public protocol EnumCollection: Hashable {}
extension EnumCollection {
public static func allValues() -> [Self] {
typealias S = Self
let retVal = AnySequence { () -> AnyIterator<S> in
var raw = 0
return AnyIterator {
let current = withUnsafePointer(to: &raw) {
$0.withMemoryRebound(to: S.self, capacity: 1) { $0.pointee }
}
guard current.hashValue == raw else { return nil }
raw += 1
return current
}
}
return [S](retVal)
}
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++) {
...
}
原则上,如果你不为enum的大小写使用原始值赋值,这样做是可能的:
enum RankEnum: Int {
case Ace
case One
case Two
}
class RankEnumGenerator: Generator {
var i = 0
typealias Element = RankEnum
func next() -> Element? {
let r = RankEnum.fromRaw(i)
i += 1
return r
}
}
extension RankEnum {
static func enumerate() -> SequenceOf<RankEnum> {
return SequenceOf<RankEnum>({ RankEnumGenerator() })
}
}
for r in RankEnum.enumerate() {
println("\(r.toRaw())")
}
(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]
}
}
}
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
}