Swift 枚举可以将 function/closure 作为原始值吗？

Question

我知道枚举可以有一个闭包作为关联的值，例如：

enum SomeEnum {
    case closureOne (String, Double -> Double)
    case closureTwo (String, (Double, Double) -> Double)
}

但是，枚举可以作为 raw 值的闭包吗？例如，这样的事情行得通吗？

enum someEnum: () -> Void {
    case closureOne = doSomething
    case closureTwo  = doSomethingElse
}

哪里

let doSomething = {
    // Do something here.
}

let doSomethingElse {
    // Do something else here.
}

Answer 1

它不是那么简单，但您可以使用 OptionSet，请参阅 this page:

Unlike enumerations, option sets provide a nonfailable init(rawValue:) initializer to convert from a raw value, because option sets don’t have an enumerated list of all possible cases. Option set values have a one-to-one correspondence with their associated raw values.

可能是这样的：

func doSomething() {}
func doSomethingElse() {}

struct MyClosures: OptionSet {

    static let closureOne = MyClosures(rawValue: doSomething)
    static let closureTwo = MyClosures(rawValue: doSomethingElse)

    let rawValue: () -> Void

    init(rawValue: @escaping () -> Void) {
        self.rawValue = rawValue
    }

    init() {
        rawValue = {}
    }

    mutating func formUnion(_ other: __owned MyClosures) {
        // whatever makes sense for your case
    }

    mutating func formIntersection(_ other: MyClosures) {
        // whatever makes sense for your case
    }

    mutating func formSymmetricDifference(_ other: __owned MyClosures) {
        // whatever makes sense for your case
    }

    static func == (lhs: MyClosures, rhs: MyClosures) -> Bool {
        // whatever makes sense for your case
        return false
    }
}

因此您可以将其用作：

let myClosures: MyClosures = [ .closureOne, .closureTwo ]

不过看看你在评论中的解释：

So I'm trying to find the most efficient way to run a function given the state of a variable.

我认为您真正想要的是某种状态机。提供了一些示例 here and here

Swift 枚举可以将 function/closure 作为原始值吗？

Can a Swift enum have a function/closure as a raw value?

enums

closures

enumeration

swift