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
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
}
其他回答
在处理Swift 2.0时,以下是我的建议:
我已经将原始类型添加到Suit enum
enum Suit: Int {
然后:
struct Card {
var rank: Rank
var suit: Suit
func fullDeck()-> [Card] {
var deck = [Card]()
for i in Rank.Ace.rawValue...Rank.King.rawValue {
for j in Suit.Spades.rawValue...Suit.Clubs.rawValue {
deck.append(Card(rank:Rank(rawValue: i)! , suit: Suit(rawValue: j)!))
}
}
return deck
}
}
其他的解决方法都是可行的,但它们都假设了可能的等级和花色的数量,或者第一和最后的等级是什么。的确,在可预见的未来,一副纸牌的布局可能不会有太大变化。然而,一般来说,编写尽可能少假设的代码会更简洁。我的解决方案:
我已经在Suit枚举中添加了一个原始类型,所以我可以使用Suit(rawValue:)来访问Suit案例:
enum Suit: Int {
case Spades = 1
case Hearts, Diamonds, Clubs
func simpleDescription() -> String {
switch self {
case .Spades:
return "spades"
case .Hearts:
return "hearts"
case .Diamonds:
return "diamonds"
case .Clubs:
return "clubs"
}
}
func color() -> String {
switch self {
case .Spades:
return "black"
case .Clubs:
return "black"
case .Diamonds:
return "red"
case .Hearts:
return "red"
}
}
}
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.rawValue)
}
}
}
在Card的createDeck()方法实现的下面。init(rawValue:)是一个可失败的初始化式,返回一个可选值。通过在两个while语句中展开并检查它的值,不需要假设Rank或Suit情况的数量:
struct Card {
var rank: Rank
var suit: Suit
func simpleDescription() -> String {
return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
}
func createDeck() -> [Card] {
var n = 1
var deck = [Card]()
while let rank = Rank(rawValue: n) {
var m = 1
while let suit = Suit(rawValue: m) {
deck.append(Card(rank: rank, suit: suit))
m += 1
}
n += 1
}
return deck
}
}
下面是如何调用createDeck方法:
let card = Card(rank: Rank.Ace, suit: Suit.Clubs)
let deck = card.createDeck()
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)
}
与@Kametrixom的答案在这里,我相信返回一个数组将比返回AnySequence更好,因为你可以访问所有数组的好东西,如计数等。
以下是改写后的内容:
public protocol EnumCollection : Hashable {}
extension EnumCollection {
public static func allValues() -> [Self] {
typealias S = Self
let retVal = AnySequence { () -> AnyGenerator<S> in
var raw = 0
return AnyGenerator {
let current : Self = withUnsafePointer(&raw) { UnsafePointer($0).memory }
guard current.hashValue == raw else { return nil }
raw += 1
return current
}
}
return [S](retVal)
}
}
我添加了函数count(),并迭代值:
public enum MetricType: Int {
case mvps = 0
case allNBA = 1
case championshipRings = 2
case finalAppearances = 3
case gamesPlayed = 4
case ppg = 5
static func count() -> Int {
return (ppg.rawValue) + 1
}
static var allValues: [MetricType] {
var array: [MetricType] = Array()
var item : MetricType = MetricType.mvps
while item.rawValue < MetricType.count() {
array.append(item)
item = MetricType(rawValue: (item.rawValue + 1))!
}
return array
}
}