匿名函数的参数类型 [英] Argument type of anonymous function

问题描述

我在使用此代码时遇到了一些问题.它应该是一个带有元素 BinaryOperations 和 UnaryOperations 的 OperationTree.方法 eval 进行评估并在地图中查找变量.

I'm having some trouble with this code. It's supposed to be an OperationTree with Elements BinaryOperations and UnaryOperations. The method eval does the evaluation and looks up the variables in a map.

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！

推荐答案

您忘记实际匹配某些内容...

You have forgotten to actually match something...

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 {

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