Scala继承参数化构造函数

Question

我有一个带有几个可选参数的抽象基类:

abstract case class Hypothesis(
    requirement: Boolean = false,
    onlyDays:   Seq[Int] = Nil,
    …
) extends Something {…}

我是否真的需要在顶部使用附加关键字override val明确重复所有参数‽

case class SomeHypothesis(
    anotherArg: SomeType,
    override val requirement: Boolean = false,
    override val onlyDays:   Seq[Int] = Nil,
    …
) extends Hypothesis(
    requirement,
    onlyDays,
    …
) {…}

或者是否有类似的语法

case class SomeHypothesis(anotherArg: SomeType, **) extends Hypothesis(**) {…}

我甚至不需要anotherArg,只是将所有关键字args传递给超级构造函数的方法.

---

我真的很喜欢Scala关于构造函数的想法,但是如果没有这个的语法,我会很失望的:(

Answer 1

您可以在继承的类中使用虚拟名称:

case class SomeHypothesis(anotherArg: SomeType, rq: Boolean = false, odays: Seq[Int] = Nil)
extends Hypothesis(rq, odays)

但你必须重复默认值.没有必要覆盖val.

编辑:

请注意,您的抽象类不应该是案例类.现在不推荐使用扩展案例类.您应该为抽象类使用提取器:

abstract class SomeHypothesis(val request: Boolean)

object SomeHypothesis {
  def unapply(o: Any): Option[Boolean] = o match {
    case sh: SomeHypothesis => Some(sh.request)
    case _ => None
  }
}

Answer 2

在我看来,默认值的策略不属于基类,而应该继续使用具体的类.我会改为做以下事情:

trait Hypothesis {
  def requirement: Boolean
  def onlyDays: Seq[Int]
  /* other common attributes as necessary */
}

case class SomeHypothesis(anotherArg: SomeType,
                          requirement: Boolean = false,
                          onlyDays: Seq[Int] = Nil)
  extends Hypothesis

案例类字段SomeHypothesis将满足假设特征的要求.

正如其他人所说,你仍然可以在公共部分使用提取器进行模式匹配:

object Hypothesis {
  def unapply(h: Hypothesis): (Boolean, Seq[Int]) = (h.requirement, h.onlyDays)
}