匿名函数的参数类型 [英] Argument type of anonymous function
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问题描述
我在使用此代码时遇到了麻烦. 它应该是一个带有元素BinaryOperations和UnaryOperations的OperationTree. eval方法进行求值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
您忘记了实际匹配的东西...
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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