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
该解决方案在可读性和可维护性之间取得了适当的平衡。
struct Card {
// ...
static func deck() -> Card[] {
var deck = Card[]()
for rank in Rank.Ace.toRaw()...Rank.King.toRaw() {
for suit in [Suit.Spades, .Hearts, .Clubs, .Diamonds] {
let card = Card(rank: Rank.fromRaw(rank)!, suit: suit)
deck.append(card)
}
}
return deck
}
}
let deck = Card.deck()
其他回答
更新代码:Swift 4.2/Swift 5
enum Suit: String, CaseIterable {
case spades = "♠"
case hearts = "♥"
case diamonds = "♦"
case clubs = "♣"
}
按问题访问输出:
for suitKey in Suit.allCases {
print(suitKey.rawValue)
}
输出:
♠
♥
♦
♣
CaseIterable:提供其所有值的集合。 符合CaseIterable协议的类型通常是没有关联值的枚举。当使用CaseIterable类型时,您可以通过使用该类型的allCases属性访问该类型的所有案例的集合。
对于访问case,我们使用。allcases。更多信息请点击https://developer.apple.com/documentation/swift/caseiterable
在Swift 3中,当底层枚举有rawValue时,你可以实现Strideable协议。优点是不像其他建议那样创建值数组,并且标准的Swift“for in”循环工作,这是一个很好的语法。
// "Int" to get rawValue, and Strideable so we can iterate
enum MyColorEnum: Int, Strideable {
case Red
case Green
case Blue
case Black
// required by Strideable
typealias Stride = Int
func advanced(by n:Stride) -> MyColorEnum {
var next = self.rawValue + n
if next > MyColorEnum.Black.rawValue {
next = MyColorEnum.Black.rawValue
}
return MyColorEnum(rawValue: next)!
}
func distance(to other: MyColorEnum) -> Int {
return other.rawValue - self.rawValue
}
// just for printing
func simpleDescription() -> String {
switch self {
case .Red: return "Red"
case .Green: return "Green"
case .Blue: return "Blue"
case .Black: return "Black"
}
}
}
// this is how you use it:
for i in MyColorEnum.Red ... MyColorEnum.Black {
print("ENUM: \(i)")
}
以下是我的建议。这不是完全令人满意的(我对Swift和OOP很陌生!),但也许有人可以改进它。这个想法是让每个枚举提供自己的范围信息作为.first和.last属性。它只向每个枚举添加了两行代码:仍然有点硬编码,但至少它没有复制整个集合。它确实需要将Suit enum修改为Int类型,就像Rank enum一样,而不是无类型的。
而不是重复整个解决方案,下面是我添加到。在case语句之后的某个地方(Suit enum类似):
var first: Int { return Ace.toRaw() }
var last: Int { return King.toRaw() }
以及我用来将deck构建为String数组的循环。(问题定义没有说明牌组是如何构造的。)
func createDeck() -> [String] {
var deck: [String] = []
var card: String
for r in Rank.Ace.first...Rank.Ace.last {
for s in Suit.Hearts.first...Suit.Hearts.last {
card = Rank.simpleDescription( Rank.fromRaw(r)!)() + " of " + Suit.simpleDescription( Suit.fromRaw(s)!)()
deck.append( card)
}
}
return deck
}
这并不令人满意,因为属性与元素而不是enum相关联。但它确实为“for”循环增加了清晰度。我希望它是Rank。而不是Rank.Ace.first。它适用于任何元素,但很难看。有人能演示一下如何将其提升到enum级别吗?
为了使它工作,我从Card结构中提取了createDeck方法。我不知道如何从该结构返回一个[String]数组,这似乎是一个糟糕的地方,把这样的方法无论如何。
enum Rank: Int {
...
static let ranks = (Rank.Ace.rawValue ... Rank.King.rawValue).map{Rank(rawValue: $0)! }
}
enum Suit {
...
static let suits = [Spades, Hearts, Diamonds, Clubs]
}
struct Card {
...
static func fullDesk() -> [Card] {
var desk: [Card] = []
for suit in Suit.suits {
for rank in Rank.ranks {
desk.append(Card(rank: rank,suit: suit))
}
}
return desk
}
}
这个怎么样?
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
}
}