Scala继承参数化构造函数 [英] Scala inherit parameterized constructor

问题描述

我有一个带有几个可选参数的抽象基类：

 抽象案例类假设（
要求：布尔值= false，
 onlyDays：Seq [Int] = Nil，
 ... 
）延伸Something {...} 
  

我是否真的需要使用附加关键字覆盖val 在上显式重复所有参数

 案例类SomeHypothesis（
 anotherArg：SomeType，
覆盖val要求：Boolean = false，
覆盖val onlyDays：Seq [Int] = Nil，
 ... 
）扩展Hypothesis（
要求，
 onlyDays，
 ... 
）{...} 
  

或者是否存在类似

<$ p $的语法p> 案例类SomeHypothesis（anotherArg：SomeType，**）扩展假设（**）{...}


我甚至不需要 anotherArg ，只是一种将所有关键字args传递给super con的方法结构。

---

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

解决方案

你可以在继承的类中使用虚拟名称：

 案例类SomeHypothesis（anotherArg：SomeType，rq：Boolean = false，odays：Seq [Int] = Nil）
扩展假设（rq，odays） 
  

但你必须重复默认值。无需覆盖 val 。

编辑：

请注意，您的抽象类不应该是案例类。扩展案例类现在已弃用。您应该使用提取器代替抽象类：

  abstract class SomeHypothesis（val request：Boolean）
 
 object SomeHypothesis {
 def unapply（o：Any）：Option [Boolean] = o匹配{
 case sh：SomeHypothesis =>一些（sh.request）
 case _ =>无
} 
} 
  

I have an abstract base class with several optional parameters:

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

Do i really need to explicitly repeat all parameters with the additional keywords override val on top‽

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

Or is there a syntax like

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

I don’t even need anotherArg, just a way to pass all keyword args to the super constructor.

---

I really like Scala’s idea about constructors, but if there isn’t a syntax for that one, I’ll be disappoint :(

解决方案

You can just use a dummy name in the inherited class:

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

but you do have to repeat the default values. There is no need to override a val.

EDIT:

Note that your abstract class should not be a case class. Extending case classes is now deprecated. You should use an extractor for you abstract class instead:

abstract class SomeHypothesis(val request: Boolean)

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

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