使用特征中定义的 unapply

Question

我有一个特点

trait A {
    def doSomething(a: Seq[Int]): Seq[String] = {
        a.map {
            case AA(s) => s // want to use unapply defined in trait (this(AA) not allowed)
            case _ => "idc"
        }
    }
    
    def unapply(a: Int): Option[String] = getString(a)

    
    def getString(a: Int): Option[String] = {
        a match {
            case 1 => Some("one")
            case 2 => Some("two")
            case _ => None
        }
    }
}

object AA extends A
object AA2 extends A {
    override def getString(a: Int): Option[String] = {
        super.getString(a).orElse{
            a match {
                case 3 => Some("three")
                case 4 => Some("four")
                case _ => None
            }
        }
    } 
}

object MyClass {
    def main(args: Array[String]) {
        println(AA.doSomething(Seq(1,2,3,4,5))); // Output: List(one, two, idc, idc, idc)
        println(AA2.doSomething(Seq(1,2,3,4,5))); // Expect Output: List(one, two, three, four, idc) but get List(one, two, idc, idc, idc)
    }
}

这里的问题是我不能在不创建提取器对象的情况下使用特征中定义的 unapply。

我想使用此特性覆盖不同对象中的 getString 方法。

Answer 1

我使用的一个解决方案是

trait A {
    def doSomething(a: Seq[Int]): Seq[String] = {
        a.map {
            case Extractor(s) => s
            case _ => "idc"
        }
    }
    
    object Extractor {
        def unapply(a: Int): Option[String] = getString(a)
    }

    
    def getString(a: Int): Option[String] = {
        a match {
            case 1 => Some("one")
            case 2 => Some("two")
            case _ => None
        }
    }
}

object AA extends A
object AA2 extends A {
    override def getString(a: Int): Option[String] = {
        super.getString(a).orElse{
            a match {
                case 3 => Some("three")
                case 4 => Some("four")
                case _ => None
            }
        }
    } 
}

object MyClass {
    def main(args: Array[String]) {
        println(AA.doSomething(Seq(1,2,3,4,5))); // Output: List(one, two, idc, idc, idc)
        println(AA2.doSomething(Seq(1,2,3,4,5))); // Output: List(one, two, three, four, idc)
    }
}

Answer 2

您可以使用自我类型来引用自己。

trait A { self =>
  final def doSomething(a: Seq[Int]): Seq[String] =
    a.map {
      case self(s) => s
      case _       => "idc"
    }

  final def unapply(a: Int): Option[String] =
    getString(a)

  def getString(a: Int): Option[String] =
    a match {
      case 1 => Some("one")
      case 2 => Some("two")
      case _ => None
    }
}

按预期工作。

代码运行here.

使用特征中定义的 unapply

Using unapply defined in traits

scala

pattern-matching

unapply