scala – 如何将ADT列表分类为其变体?
发布时间:2020-12-16 18:32:10 所属栏目:安全 来源:网络整理
导读:是否有可能以某种方式将解决方案扩展为总和类型? sealed trait Groupcase class A extends Groupcase class B extends Groupcase class C extends Groupdef divide(l : List[Group]): //Something from what I can extract List[A],List[B] and List[C] 解
是否有可能以某种方式将解决方案扩展为总和类型?
sealed trait Group case class A extends Group case class B extends Group case class C extends Group def divide(l : List[Group]): //Something from what I can extract List[A],List[B] and List[C] 解决方法
也许你可以尝试改进这个答案.这可能无法解决您的问题,因为很难知道给定类型的任意子类型(组类型可能具有任意数量的子类型).在Either的情况下,很容易预测它的子类型为Right或Left.
sealed trait Group case class A(name:String) extends Group case class B(name:String) extends Group case class C(name:String) extends Group val list = List( A("a1"),A("a2"),A("a3"),A("a4"),B("b1"),B("b2"),B("b3"),B("b4"),C("c1"),C("c2"),C("c3"),C("c4") ) def divide( list: List[Group],aList : List[A],bList: List[B],cList: List[C] ): (List[A],List[B],List[C]) = { list match { case Nil => (aList,bList,cList) case head :: tail => head match { case a : A => divide(tail,aList.:+(a),cList) case b : B => divide(tail,aList,bList.:+(b),cList) case c : C => divide(tail,cList.:+(c)) } } } divide(list,List.empty[A],List.empty[B],List.empty[C]) //res1: (List[A],List[C]) = (List(A(a1),A(a2),A(a3),A(a4)),List(B(b1),B(b2),B(b3),B(b4)),List(C(c1),C(c2),C(c3),C(c4))) 希望这对你有所帮助. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |