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
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()
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
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())")
}
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)
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
}
}