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
对不起,我的回答是具体到我如何在我需要做的事情中使用这篇文章。对于那些无意中遇到这个问题的人,寻找一种方法在枚举中找到一个case,这是一种方法(Swift 2新增):
编辑:小写驼峰现在是Swift 3 enum值的标准
// From apple docs: If the raw-value type is specified as String and you don’t assign values to the cases explicitly, each unassigned case is implicitly assigned a string with the same text as the name of that case.
enum Theme: String
{
case white, blue, green, lavender, grey
}
func loadTheme(theme: String)
{
// this checks the string against the raw value of each enum case (note that the check could result in a nil value, since it's an optional, which is why we introduce the if/let block
if let testTheme = Theme(rawValue: theme)
{
// testTheme is guaranteed to have an enum value at this point
self.someOtherFunction(testTheme)
}
}
对于那些对枚举感到疑惑的人来说,本页上给出的答案包括一个包含所有枚举值的数组的静态var/let是正确的。最新的苹果tvOS示例代码包含了完全相同的技术。
也就是说,他们应该在语言中构建一个更方便的机制(苹果,你在听吗?)
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
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++) {
...
}
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())
}
}
}
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, .Clubs: return "black"
case .Hearts, .Diamonds: return "red"
}
}
}
struct Card {
var rank: Rank
var suit: Suit
func simpleDescription() -> String {
return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
}
static func createPokers() -> Card[] {
let ranks = Array(Rank.Ace.toRaw()...Rank.King.toRaw())
let suits = Array(Suit.Spades.toRaw()...Suit.Clubs.toRaw())
let cards = suits.reduce(Card[]()) { (tempCards, suit) in
tempCards + ranks.map { rank in
Card(rank: Rank.fromRaw(rank)!, suit: Suit.fromRaw(suit)!)
}
}
return cards
}
}
有一种聪明的方法,尽管令人沮丧,但它说明了两种不同类型的枚举之间的区别。
试试这个:
func makeDeck() -> Card[] {
var deck: Card[] = []
var suits: Suit[] = [.Hearts, .Diamonds, .Clubs, .Spades]
for i in 1...13 {
for suit in suits {
deck += Card(rank: Rank.fromRaw(i)!, suit: suit)
}
}
return deck
}
交易是,一个由数字(原始值)支持的枚举是隐式显式有序的,而一个没有数字支持的枚举是显式隐式无序的。
例如,当我们给枚举值数字时,语言足够狡猾,可以找出数字的顺序。 另一方面,如果我们不给它任何顺序,当我们尝试迭代这些值时,语言就会举起双手说:“是的,但你想先执行哪个??”
其他可以做到这一点(迭代无序枚举)的语言可能是相同的语言,其中所有内容实际上都是一个地图或字典,你可以迭代地图的键,无论是否有任何逻辑顺序。
诀窍是给它提供一些显式排序的东西,在这个例子中,suit的实例在数组中按照我们想要的顺序。一旦你这么说,霉霉就会说“你为什么不一开始就这么说呢?”
另一个简写技巧是在fromRaw函数上使用强制操作符。这说明了关于枚举的另一个“陷阱”,即可能传入的值的范围通常大于枚举的范围。例如,如果我们说Rank.fromRaw(60),就不会返回值,所以我们使用了语言的可选特性,在我们开始使用可选特性的地方,很快就会出现强制。(或者交替if let结构,这对我来说仍然有点奇怪)
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
}