如何调用 Scala 抽象类型的构造函数? [英] How can I invoke the constructor of a Scala abstract type?

问题描述

我正在尝试弄清楚如何为 Scala 抽象类型调用构造函数:

I'm trying to figure out how to invoke a constructor for a Scala abstract type:

class Journey(val length: Int)
class PlaneJourney(length: Int) extends Journey(length)
class BoatJourney(length: Int) extends Journey(length)

class Port[J <: Journey] {
  def startJourney: J = {
    new J(23) // error: class type required but J found
  }
}

这甚至可行吗?我熟悉 Scala 清单 但我不清楚他们如何在这里提供帮助.同样，我无法弄清楚如何对伴随对象的 apply() 构造函数执行相同的操作:

Is this even feasible? I'm familiar with Scala manifests but I'm not clear how they could help here. Likewise I can't figure out how to do the same with a companion object's apply() constructor:

object Journey { def apply() = new Journey(0) }
object PlaneJourney { def apply() = new PlaneJourney(0) }
object BoatJourney { def apply() = new BoatJourney(0) }

class Port[J <: Journey] {
  def startJourney: J = {
    J() // error: not found: value J
  }
}

感谢收到任何想法！

推荐答案

没有直接的方法来调用构造函数或访问仅给定类型的伴随对象.一种解决方案是使用类型类来构造给定类型的默认实例.

There is no direct way to invoke the constructor or access the companion object given only a type. One solution would be to use a type class that constructs a default instance of the given type.

trait Default[A] { def default: A }

class Journey(val length: Int)
object Journey {
  // Provide the implicit in the companion
  implicit def default: Default[Journey] = new Default[Journey] {
    def default = new Journey(0)
  }
}

class Port[J <: Journey : Default] {
  // use the Default[J] instance to create the instance
  def startJourney: J = implicitly[Default[J]].default
}

您需要向支持创建默认实例的类的所有伴随对象添加隐式 Default 定义.

You will need to add an implicit Default definition to all companion objects of classes that support creation of a default instance.

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