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
这篇文章是相关的https://www.swift-studies.com/blog/2014/6/10/enumerating-enums-in-swift
基本上,提议的解决方案是
enum ProductCategory : String {
case Washers = "washers", Dryers = "dryers", Toasters = "toasters"
static let allValues = [Washers, Dryers, Toasters]
}
for category in ProductCategory.allValues{
//Do something
}
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
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
}
}
这篇文章是相关的https://www.swift-studies.com/blog/2014/6/10/enumerating-enums-in-swift
基本上,提议的解决方案是
enum ProductCategory : String {
case Washers = "washers", Dryers = "dryers", Toasters = "toasters"
static let allValues = [Washers, Dryers, Toasters]
}
for category in ProductCategory.allValues{
//Do something
}
你可以试着像这样列举
enum Planet: String {
case Mercury
case Venus
case Earth
case Mars
static var enumerate: [Planet] {
var a: [Planet] = []
switch Planet.Mercury {
case .Mercury: a.append(.Mercury); fallthrough
case .Venus: a.append(.Venus); fallthrough
case .Earth: a.append(.Earth); fallthrough
case .Mars: a.append(.Mars)
}
return a
}
}
Planet.enumerate // [Mercury, Venus, Earth, Mars]
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