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
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
}
其他回答
更新到Swift 2.2+
func iterateEnum<T: Hashable>(_: T.Type) -> AnyGenerator<T> {
var i = 0
return AnyGenerator {
let next = withUnsafePointer(&i) {
UnsafePointer<T>($0).memory
}
if next.hashValue == i {
i += 1
return next
} else {
return nil
}
}
}
它更新了Swift 2.2表单@Kametrixom的答案
Swift 3.0+(非常感谢@Philip)
func iterateEnum<T: Hashable>(_: T.Type) -> AnyIterator<T> {
var i = 0
return AnyIterator {
let next = withUnsafePointer(&i) {
UnsafePointer<T>($0).pointee
}
if next.hashValue == i {
i += 1
return next
} else {
return nil
}
}
}
下面是我用来迭代枚举和从一个枚举提供多个值类型的方法
enum IterateEnum: Int {
case Zero
case One
case Two
case Three
case Four
case Five
case Six
case Seven
//tuple allows multiple values to be derived from the enum case, and
//since it is using a switch with no default, if a new case is added,
//a compiler error will be returned if it doesn't have a value tuple set
var value: (french: String, spanish: String, japanese: String) {
switch self {
case .Zero: return (french: "zéro", spanish: "cero", japanese: "nuru")
case .One: return (french: "un", spanish: "uno", japanese: "ichi")
case .Two: return (french: "deux", spanish: "dos", japanese: "ni")
case .Three: return (french: "trois", spanish: "tres", japanese: "san")
case .Four: return (french: "quatre", spanish: "cuatro", japanese: "shi")
case .Five: return (french: "cinq", spanish: "cinco", japanese: "go")
case .Six: return (french: "six", spanish: "seis", japanese: "roku")
case .Seven: return (french: "sept", spanish: "siete", japanese: "shichi")
}
}
//Used to iterate enum or otherwise access enum case by index order.
//Iterate by looping until it returns nil
static func item(index: Int) -> IterateEnum? {
return IterateEnum.init(rawValue: index)
}
static func numberFromSpanish(number: String) -> IterateEnum? {
return findItem { $0.value.spanish == number }
}
//use block to test value property to retrieve the enum case
static func findItem(predicate: ((_: IterateEnum) -> Bool)) -> IterateEnum? {
var enumIndex: Int = -1
var enumCase: IterateEnum?
//Iterate until item returns nil
repeat {
enumIndex += 1
enumCase = IterateEnum.item(index: enumIndex)
if let eCase = enumCase {
if predicate(eCase) {
return eCase
}
}
} while enumCase != nil
return nil
}
}
var enumIndex: Int = -1
var enumCase: IterateEnum?
// Iterate until item returns nil
repeat {
enumIndex += 1
enumCase = IterateEnum.item(index: enumIndex)
if let eCase = enumCase {
print("The number \(eCase) in french: \(eCase.value.french), spanish: \(eCase.value.spanish), japanese: \(eCase.value.japanese)")
}
} while enumCase != nil
print("Total of \(enumIndex) cases")
let number = IterateEnum.numberFromSpanish(number: "siete")
print("siete in japanese: \((number?.value.japanese ?? "Unknown"))")
输出如下:
法语中的数字Zero: zéro,西班牙语中的数字cero,日语中的数字nuru 数字一在法语中是un,西班牙语中是uno,日语中是ichi 法语中的数字2是deux,西班牙语中的数字2是dos,日语中的数字2是ni 法语中的“三”是“trois”,西班牙语中的“tres”,日语中的“san” 法语中的“四”是quatre,西班牙语中的“四”是cuatro,日语中的“四”是shi 数字五在法语中是cinq,西班牙语中是cinco,日语中是go 数字6在法语中是Six,西班牙语是seis,日语是roku 法语中的数字“七”是“sept”,西班牙语中的“siete”,日语中的“shichi”
共8例
Siete在日语中的意思是:shichi
更新
我最近创建了一个协议来处理枚举。该协议需要一个Int原始值的enum:
protocol EnumIteration {
//Used to iterate enum or otherwise access enum case by index order. Iterate by looping until it returns nil
static func item(index:Int) -> Self?
static func iterate(item:((index:Int, enumCase:Self)->()), completion:(()->())?) {
static func findItem(predicate:((enumCase:Self)->Bool)) -> Self?
static func count() -> Int
}
extension EnumIteration where Self: RawRepresentable, Self.RawValue == Int {
//Used to iterate enum or otherwise access enum case by index order. Iterate by looping until it returns nil
static func item(index:Int) -> Self? {
return Self.init(rawValue: index)
}
static func iterate(item:((index:Int, enumCase:Self)->()), completion:(()->())?) {
var enumIndex:Int = -1
var enumCase:Self?
//Iterate until item returns nil
repeat {
enumIndex += 1
enumCase = Self.item(enumIndex)
if let eCase = enumCase {
item(index: enumIndex, enumCase: eCase)
}
} while enumCase != nil
completion?()
}
static func findItem(predicate:((enumCase:Self)->Bool)) -> Self? {
var enumIndex:Int = -1
var enumCase:Self?
//Iterate until item returns nil
repeat {
enumIndex += 1
enumCase = Self.item(enumIndex)
if let eCase = enumCase {
if predicate(enumCase:eCase) {
return eCase
}
}
} while enumCase != nil
return nil
}
static func count() -> Int {
var enumIndex:Int = -1
var enumCase:Self?
//Iterate until item returns nil
repeat {
enumIndex += 1
enumCase = Self.item(enumIndex)
} while enumCase != nil
//last enumIndex (when enumCase == nil) is equal to the enum count
return enumIndex
}
}
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)")
}
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。
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++)
}
}
)
}
}