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++)
}
}
)
}
}
其他回答
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
}
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
}
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
}
与@Kametrixom的答案在这里,我相信返回一个数组将比返回AnySequence更好,因为你可以访问所有数组的好东西,如计数等。
以下是改写后的内容:
public protocol EnumCollection : Hashable {}
extension EnumCollection {
public static func allValues() -> [Self] {
typealias S = Self
let retVal = AnySequence { () -> AnyGenerator<S> in
var raw = 0
return AnyGenerator {
let current : Self = withUnsafePointer(&raw) { UnsafePointer($0).memory }
guard current.hashValue == raw else { return nil }
raw += 1
return current
}
}
return [S](retVal)
}
}
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)
}
}
//...
}