如何在 Scala 的编译时验证类型是否存在

Question

我有以下特质和职业：

sealed trait Signal

sealed trait Description[T]

final case class S1(name: String) extends Signal

final case class D1(name: String) extends Description[S1]

我试图实现的是，任何想要添加 Signal 的人都可以（在编译时）创建描述。

我不想更改Description 的签名，但肯定不会更改Signal 的签名

我将编译器设置为在警告时失败，因此我可以利用我的 ADT 已密封这一事实。

我的想法是有这样一个“编译守卫”：

def compilationGuard[S <: Signal](s: S): Description[S] = s match { case S1(name) => D1(name) }

但我收到以下错误：

<console>:17: error: type mismatch;
 found   : D1
 required: Description[S]
       def compilationGuard[S <: Signal](s: S): Description[S] = s match { case S1(name) => D1(name) }
                                                                                              ^

Answer 1

def compilationGuard[S <: Signal](s: S): Description[S] = s match { case S1(name) => D1(name) }

由于与

相同的原因无法编译

def returnItself[S <: Signal](s: S): S = s match { case S1(name) => S1(name) }

这里详细解释原因：

Why can't I return a concrete subtype of A if a generic subtype of A is declared as return parameter?

Type mismatch on abstract type used in pattern matching

如果您不想将 Description 逻辑与 ADT 混合或手动定义类型类的实例，例如 SignalMapper，您可以使用 Shapeless

import shapeless.ops.coproduct.Mapper
import shapeless.{:+:, CNil, Coproduct, Generic, Poly1}

def compilationGuard[C <: Coproduct]()(implicit
  gen: Generic.Aux[Signal, C],
  mapper: Mapper[uniqueDescriptionPoly.type, C]
) = null

object uniqueDescriptionPoly extends Poly1 {
  implicit def cse[S <: Signal, C1 <: Coproduct](implicit
    gen1: Generic.Aux[Description[S], C1],
    ev: C1 <:< (_ :+: CNil)
  ): Case.Aux[S, Null] = null
}

compilationGuard()

测试：

final case class S1(name: String) extends Signal
final case class S2(name: String) extends Signal
final case class D1(name: String) extends Description[S1] 
// doesn't compile

final case class S1(name: String) extends Signal
final case class S2(name: String) extends Signal
final case class D1(name: String) extends Description[S1]
final case class D2(name: String) extends Description[S1]
// doesn't compile

final case class S1(name: String) extends Signal
final case class S2(name: String) extends Signal
final case class D1(name: String) extends Description[S1]
final case class D2(name: String) extends Description[S2]
// compiles

Answer 2

您的程序失败了，因为它无法证明类型 S 是 S1。

你可以引入类型类来代替模式匹配，它会在编译时进行映射：

trait SignalMapper[S] { //typeclass handling of mapping S to D
  type D <: Description[S]

  def map(signal: S): D
}

//instance of typeclass SignalMapper for S1
//if you'd put it in a companion object of S1, it would be always in scope
object S1 { 

  implicit val mapperS1: SignalMapper[S1] = new SignalMapper[S1] {
    type D = D1

    def map(signal: S1) = D1(signal.name)
  }

}

然后你可以重写compileGuard为：

def compilationGuard[S <: Signal](s: S)(implicit mapper: SignalMapper[S]): Description[S] = mapper.map(s)

Scastie

Answer 3

我试图实现的是，任何想要添加 Signal 的人都将拥有（在编译时创建描述。

这样做的简单方法是使其成为Signal 的一部分：

sealed trait Signal[S <: Signal[S, D], D <: Description[S]] {
  // optionally
  def description: D
}

final case class S1(name: String) extends Signal[S1, D1] {
  def description = D1(name)
}

或

sealed trait Signal[S <: Signal[S]] {
  type Descr <: Description[S]
  // optionally
  def description: Descr
}

final case class S1(name: String) extends Signal[S1] {
  type Descr = D1
  def description = D1(name)
}

当然也不是很简单

sealed trait Signal[S <: Signal] {
  def description: Description[S]
}

取决于您的要求。