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更高种类的类型作为类型参数

higher-kinded-types kotlin

Tobia · 技术社区 · 7 年前

我有一个通用的密封类,用于表示单个值或成对值(在某个事件之前和之后拆分):

sealed class Splittable<T>

data class Single<T>(val single: T) : Splittable<T>()

data class Split<T>(val before: T, 
                    val after : T) : Splittable<T>()

Splittable ,因此类的属性必须 要么全部是单一的,要么全部是分裂的。 我以为这样就行了:

data class Totals<SInt : Splittable<Int>>(
    val people : SInt,
    val things : SInt
)

val t1 = Totals(
    people = Single(3),
    things = Single(4)
)

val t2 = Totals(
    people = Split(3, 30),
    things = Split(4, 40)
)

但我错了,因为它允许无效的组合:

val WRONG = Totals(
    people = Single(3),
    things = Split(4, 40)
)

此外,如果我的类有多个基本类型,例如两个 Int 和 Boolean ? 如何编写通用签名?

data class TotalsMore<S : Splittable>(
    val people : S<Int>,
    val things : S<Int>,
    val happy  : S<Boolean>
)

val m1 = TotalsMore(
    people = Single(3),
    things = Single(4),
    happy  = Single(true)
)

val m2 = TotalsMore(
    people = Split(3, 30),
    things = Split(4, 40),
    happy  = Split(true, false)
)

error: one type argument expected for class Splittable<T>
    data class TotalsMore<S : Splittable>(
                              ^
error: type arguments are not allowed for type parameters
        val people : S<Int>,
                      ^
error: type arguments are not allowed for type parameters
        val things : S<Int>,
                      ^
error: type arguments are not allowed for type parameters
        val happy  : S<Boolean>
                      ^

所以看起来我不能传递一个更高种类的类型作为类型参数。真倒霉。

我可以将两个泛型解耦:

data class TotalsMore<SInt : Splittable<Int>,
                      SBoolean: Splittable<Boolean>>(
    val people : SInt,
    val things : SInt,
    val happy  : SBoolean
)

val WRONG = TotalsMore(
    people = Single(3),
    things = Single(4),
    happy  = Split(true, false)
)

我希望这些数据类的每个对象要么由所有单个值组成,要么由所有拆分值组成,而不是混合匹配。

我能用Kotlin的类型来表达吗?

1 回复 | 直到 7 年前

Dario Seidl 7 年前

你的 Totals 类需要泛型类型参数,但您没有在示例的构造函数中指定泛型类型参数。其工作方式是使用类型推断:Kotlin编译器从其他参数中计算出泛型类型。你可以混合的原因 Single Split WRONG 例如,编译器看到这两个参数并从中推断出公共超类型。所以你实际上是在构建一个 Totals<Splittable<Int>> .

如果显式指定子类型,则无法混合:

val WRONG = Totals<Single<Int>>(
    people = Single(3),
    things = Split(4, 40) /**  Type inference failed. Expected type mismatch: inferred type is Split<Int> but Single<Int> was expected */
)

所以你要做的是接受 Splittable 可拆分的 本身作为泛型参数。

您可以通过子类的附加接口和附加的 generic constraint where -条款:

sealed class Splittable<T>

interface ConcreteSplittable

data class Single<T>(val single: T) : Splittable<T>(), ConcreteSplittable

data class Split<T>(val before: T, 
                    val after : T) : Splittable<T>(), ConcreteSplittable

data class Totals<SInt : Splittable<Int>>(
    val people : SInt,
    val things : SInt
) where SInt : ConcreteSplittable

val t1 = Totals<Single<Int>>(
    people = Single(3),
    things = Single(4)
)

val t2 = Totals(
    people = Split(3, 30),
    things = Split(4, 40)
)

val WRONG = Totals( /** Type parameter bound for SInt in constructor Totals<SInt : Splittable<Int>>(people: SInt, things: SInt) where SInt : ConcreteSplittable is not satisfied: inferred type Any is not a subtype of Splittable<Int> */
    people = Single(3),
    things = Split(4, 40)
)

不幸的是,您也不能引入第三个类型参数 S SInt 和 SBool S Splittable<Int> 或 Splittable<Bool>

data class TotalsMore<S, SInt, SBool>
(
    val people : SInt,
    val things : SInt,
    val happy  : SBool
) where S : ConcreteSplittable, 
  SInt : S, 
  SInt : Splittable<Int>, /** Type parameter cannot have any other bounds if it's bounded by another type parameter */
  SBool : S,
  SBool : Splittable<Boolean> /** Type parameter cannot have any other bounds if it's bounded by another type parameter */

您可以做的是创建“安全”类型别名,如下所示:

data class TotalsMore<SInt : Splittable<Int>, SBool : Splittable<Boolean>> (
    val people : SInt,
    val things : SInt,
    val happy  : SBool )

typealias SingleTotalsMore = TotalsMore<Single<Int>, Single<Boolean>>

typealias SplitTotalsMore = TotalsMore<Split<Int>, Split<Boolean>>

val s = SingleTotalsMore(
    people = Single(3),
    things = Single(4),
    happy  = Single(true) )

创建混合 TotalsMore