Kotlin 枚举就像在 swift 中一样

Question

我在swift中有枚举

enum Type {
    case bool(Bool)
    case int(Int)
    case array([String])
}

不明白如何将其转换为 kotlin 代码，我确实喜欢这样:

enum class AnswerSheetType {
    BOOL,
    INT,
    ARRAY
}

但是我如何将变量传递给枚举类型。例如，我想要创建方法，该方法将返回类型为变量，如下所示(swift 代码):

func marks(for id: String) -> Type {
    let answer = answers?[id]
    
    if let boolAnswer = answer as? Bool {
        return .bool(boolAnswer)
    }
    if let intAnswer = answer as? Int {
        return .int(intAnswer)
    }
    if let arrayAnswer = answer as? [String] {
        return .array(arrayAnswer)
    }
}

Answer 1

您可以使用密封的接口(interface)/类来表示这一点。

sealed interface Type {

    data class BoolType(val value: Bool) : Type
    data class IntType(val value: Int) : Type
    data class ArrayType(val value: Array<String>) : Type

    // if you have a case that doesn't have any associated values, just use an object
    // object CaseWithoutAssociatedValues: Type

}

用法:

// let someType: Type = .bool(true)
val someType: Type = Type.BoolType(true)

// just like how you can use a switch on a Swift enum, you can use a when like this too:
// This when is also exhaustive, if you specify all the implementers of Type
when (someType) {
    is Type.BoolType -> println("Bool value: ${someType.value}")
    is Type.IntType -> println("Int value: ${someType.value}")
    is Type.ArrayType -> println("Array value: ${someType.value}")
}

请注意，在每个分支中，由于智能转换，您可以访问 someType.value。这与 Swift 不同，在 Swift 中，您需要进行模式匹配以获取相关值。