scala - 匿名函数的参数类型

Question

我在使用这段代码时遇到了一些问题。它应该是一个带有元素 BinaryOperations 和 UnaryOperations 的操作树。eval 方法进行评估并在映射中查找变量。

这是代码

 1 import collection.immutable.HashMap
  2 sealed abstract class OpTree[T]{
  3 
  4   def eval(v:HashMap[Char,T]):T = {
  5     case Elem(x) => x
  6     case UnOp(f,c) => {
  7       f(c.eval(v))
  8     }
  9     case BinOp(f,l,r) => {
 10       f(l.eval(v),r.eval(v))
 11     }
 12     case Var(c) => {
 13       v.get(c)
 14     }
 15   }
 16 }
 17 //Leaf
 18 case class Elem[T](elm:T) extends OpTree[T]
 19 //Node with two sons
 20 case class UnOp[T](f:T => T, child:OpTree[T]) extends OpTree[T]
 21 //Node with one son
 22 case class BinOp[T](f:(T,T) => T, left:OpTree[T], right:OpTree[T]) extends OpTree[T]
 23 case class Var[T](val c:Char) extends OpTree[T]

编译器说:

OpTree.scala:4: error: missing parameter type for expanded function
The argument types of an anonymous function must be fully known. (SLS 8.5)
Expected type was: T
  def eval(v:HashMap[Char,T]):T = {
                                  ^
one error found

有什么建议吗？

谢谢!

Answer 1

您忘记了实际匹配某些内容...

您的代码:

def eval(v:HashMap[Char,T]):T = {

必要的代码:

def eval(v:HashMap[Char,T]):T = v match {