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
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()
其他回答
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())")
}
我使用计算属性,它返回所有值的数组(感谢这篇文章http://natecook.com/blog/2014/10/loopy-random-enum-ideas/)。但是,它也使用int原始值,但我不需要在单独的属性中重复枚举的所有成员。
Xcode 6.1在如何使用rawValue获取enum成员方面做了一点改变,所以我修正了listing。还修复了第一个rawValue错误的小错误。
enum ValidSuits: Int {
case Clubs = 0, Spades, Hearts, Diamonds
func description() -> String {
switch self {
case .Clubs:
return "♣︎"
case .Spades:
return "♠︎"
case .Diamonds:
return "♦︎"
case .Hearts:
return "♥︎"
}
}
static var allSuits: [ValidSuits] {
return Array(
SequenceOf {
() -> GeneratorOf<ValidSuits> in
var i=0
return GeneratorOf<ValidSuits> {
return ValidSuits(rawValue: i++)
}
}
)
}
}
枚举有toRaw()和fromRaw()方法。所以如果你的原始值是Int,你可以从第一个枚举迭代到最后一个枚举:
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"
}
}
}
for i in Suit.Spades.toRaw()...Suit.Clubs.toRaw() {
if let covertedSuit = Suit.fromRaw(i) {
let description = covertedSuit.simpleDescription()
}
}
一个问题是在运行simpleDescription方法之前需要测试可选值,因此我们首先将convertedSuit设置为我们的值,然后将常量设置为convertedSuit.simpleDescription()
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()
Swift 5解决方案:
enum Suit: String, CaseIterable {
case spades = "♠"
case hearts = "♥"
case diamonds = "♦"
case clubs = "♣"
}
// access cases like this:
for suitKey in Suit.allCases {
print(suitKey)
}