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
我的解决方案是声明一个包含所有枚举可能性的数组。所以for循环可以遍历所有这些。
//Function inside struct Card
static func generateFullDeck() -> [Card] {
let allRanks = [Rank.Ace, Rank.Two, Rank.Three, Rank.Four, Rank.Five, Rank.Six, Rank.Seven, Rank.Eight, Rank.Nine, Rank.Ten, Rank.Jack, Rank.Queen, Rank.King]
let allSuits = [Suit.Hearts, Suit.Diamonds, Suit.Clubs, Suit.Spades]
var myFullDeck: [Card] = []
for myRank in allRanks {
for mySuit in allSuits {
myFullDeck.append(Card(rank: myRank, suit: mySuit))
}
}
return myFullDeck
}
//actual use:
let aFullDeck = Card.generateFullDeck() //Generate the desired full deck
var allDesc: [String] = []
for aCard in aFullDeck {
println(aCard.simpleDescription()) //You'll see all the results in playground
}
其他回答
这是一个相当老的帖子,来自Swift 2.0。现在有一些更好的解决方案,使用了swift 3.0的新特性: 在Swift 3.0中迭代一个Enum
关于这个问题,有一个解决方案,它使用了Swift 4.2的一个新功能(在我写这篇编辑时还没有发布): 我如何得到一个Swift枚举的计数?
在这个帖子中有很多好的解决方案,但其中一些非常复杂。我喜欢尽可能地简化。这里有一个解决方案,可能适用于不同的需求,但我认为它在大多数情况下都很好:
enum Number: String {
case One
case Two
case Three
case Four
case EndIndex
func nextCase () -> Number
{
switch self {
case .One:
return .Two
case .Two:
return .Three
case .Three:
return .Four
case .Four:
return .EndIndex
/*
Add all additional cases above
*/
case .EndIndex:
return .EndIndex
}
}
static var allValues: [String] {
var array: [String] = Array()
var number = Number.One
while number != Number.EndIndex {
array.append(number.rawValue)
number = number.nextCase()
}
return array
}
}
迭代:
for item in Number.allValues {
print("number is: \(item)")
}
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
}
有时,您可能会处理具有底层原始整数类型的枚举类型,这种类型在整个软件开发生命周期中都会发生变化。下面是一个很适合这种情况的例子:
public class MyClassThatLoadsTexturesEtc
{
//...
// Colors used for gems and sectors.
public enum Color: Int
{
// Colors arranged in order of the spectrum.
case First = 0
case Red, Orange, Yellow, Green, Blue, Purple, Pink
// --> Add more colors here, between the first and last markers.
case Last
}
//...
public func preloadGems()
{
// Preload all gems.
for i in (Color.First.toRaw() + 1) ..< (Color.Last.toRaw())
{
let color = Color.fromRaw(i)!
loadColoredTextures(forKey: color)
}
}
//...
}
它花了我一点,而不仅仅是一个方法在结构像swift书调用,但我在枚举中设置了下一个函数。我会使用一个协议,我不知道为什么,但有秩设置为int混乱。
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())
}
}
mutating func next() -> Rank {
var rank = self
var rawrank = rank.toRaw()
var nrank: Rank = self
rawrank = rawrank + 1
if let newRank = Rank.fromRaw(rawrank) {
println("\(newRank.simpleDescription())")
nrank = newRank
} else {
return self
}
return nrank
}
}
enum Suit {
case Spades, Hearts, Diamonds, Clubs
func color() -> String {
switch self {
case .Spades, .Clubs:
return "black"
default:
return "red"
}
}
func simpleDescription() -> String {
switch self {
case .Spades:
return "spades"
case .Hearts:
return "hearts"
case .Diamonds:
return "diamonds"
case .Clubs:
return "clubs"
}
}
mutating func next() -> Suit {
switch self {
case .Spades:
return Hearts
case .Hearts:
return Diamonds
case .Diamonds:
return Clubs
case .Clubs:
return Spades
}
}
}
struct Card {
var rank: Rank
var suit: Suit
func deck() -> Card[] {
var tRank = self.rank
var tSuit = self.suit
let tcards = 52 // we start from 0
var cards: Card[] = []
for i in 0..tcards {
var card = Card(rank: tRank, suit: tSuit)
cards.append(card)
tRank = tRank.next()
tSuit = tSuit.next()
}
return cards
}
func simpleDescription() -> String {
return "The \(rank.simpleDescription()) of \(suit.simpleDescription())"
}
}
var card = Card(rank: .Ace, suit: .Spades)
var deck = card.deck()
我使用了一些常识,但这可以通过将花色乘以等级来轻松纠正(如果你没有使用标准的桥牌,你必须相应地改变枚举,如果基本上只是通过不同的枚举进行步骤)。为了节省时间,我使用了ranks rawValues,如果你愿意,你也可以为西装做同样的事情。然而,这个例子没有它,所以我决定在不改变suit rawValue的情况下找出它
我发现自己在代码中经常使用. allvalues。我终于找到了一种方法来简单地遵循Iteratable协议并拥有一个rawValues()方法。
protocol Iteratable {}
extension RawRepresentable where Self: RawRepresentable {
static func iterateEnum<T: Hashable>(_: T.Type) -> AnyIterator<T> {
var i = 0
return AnyIterator {
let next = withUnsafePointer(to: &i) {
$0.withMemoryRebound(to: T.self, capacity: 1) { $0.pointee }
}
if next.hashValue != i { return nil }
i += 1
return next
}
}
}
extension Iteratable where Self: RawRepresentable, Self: Hashable {
static func hashValues() -> AnyIterator<Self> {
return iterateEnum(self)
}
static func rawValues() -> [Self.RawValue] {
return hashValues().map({$0.rawValue})
}
}
// Example
enum Grocery: String, Iteratable {
case Kroger = "kroger"
case HEB = "h.e.b."
case Randalls = "randalls"
}
let groceryHashes = Grocery.hashValues() // AnyIterator<Grocery>
let groceryRawValues = Grocery.rawValues() // ["kroger", "h.e.b.", "randalls"]