scala正确定义抽象数据类型的空值

你想要的是树是协变的：如果X&lt ;: Y那么树[X]&lt ;:树[Y].然后,对于任何A,Nothing<：A,你得到EmptyTree.type<：Tree [A],你总是可以在需要树[A]时传递EmptyTree. 在树协变中声明A参数的语法是Tree [A];改变它,你的代码应该编译. 这是关于Scala：Be friend with covariance and contravariance中协方差和逆变的一篇好文章

更新在questioning answer之后,我实际上已经查看了Tree的构造函数,并且根据定义,您无法使Tree协变.可悲的是,编译器不会抱怨(你看,它实际上应该抱怨更多).您在Node中的操作在Seq [A]上是逆变的,因此您无法使节点协变.此时你可能会想：

Who cares about Node? I just want Tree to be covariant!

好吧,通过使其超类型树协变节点在实践中变得如此. scalac实际上应该检查协变者的所有子类型构造函数是否(或可能是)协变的.无论如何,代码显示如下：

// you need a value for EmptyTree! thus default
def evaluateTree[Z](tree: Tree[Z],default: Z): Z =
  tree match {
    case EmptyTree    => default
    case Leaf(value)  => value
    // note how you need to essentially cast here
    case Node(op: (Seq[Z] => Z),args @ _*) =>
      op(args map { branches => evaluateTree(branches,default) })
  }

trait A
trait B extends A

val notNice: Tree[A] = Node[B]({ bs: Seq[B] => bs.head },EmptyTree)
// ClassCastException!
val uhoh = evaluateTree(notNice,new A {})

更新2回到原来的问题:)我会让你的树类型不变,并有一个EmptyTree [A]()案例类;遗憾的是没有无参数值类.

sealed trait Tree[A]
case class EmptyTree[A]() extends Tree[A]
case class Leaf[A](value: A) extends Tree[A]
// I wouldn't use varargs here,make a method for that if you want
case class Node[A](op: Seq[A] => A,branches: Tree[A]*) extends Tree[A]
// for convenience,it could be inside `Tree` companion
def emptyTree[A]: EmptyTree[A] = EmptyTree()

def evaluateTree[Z](tree: Tree[Z],default: Z): Z =
  tree match {
    case EmptyTree() =>
      default
    case Leaf(value) =>
      value
    // no need to match generic types or anything here
    case Node(op,default) })
  }

trait A
trait B extends A

// doesn't work now
// val notNice: Tree[A] = Node[B]({ bs: Seq[B] => bs.head },emptyTree)
val notNice: Tree[B] = Node[B]({ bs: Seq[B] => bs.head },emptyTree)

// doesn't compile,no class cast exception
// val uhoh = evaluateTree(notNice,new A {})