Scala 中的案例类匹配

问题描述

我想要这个匹配两个 case 类的比较函数，但它有点冗长.

I want to have this comparison function that matches two case classes, but it's a bit verbose.

叶子总是按照列表中的排序顺序.

Leafs are always in the sorted order in the List.

  abstract class CodeTree
  case class Fork(left: CodeTree, right: CodeTree, chars: List[Char], weight: Int) extends CodeTree
  case class Leaf(char: Char, weight: Int) extends CodeTree

  def sortCodeTreeFun(x: CodeTree, y: CodeTree) = {
    (x, y) match {
      case (x1: Leaf, y1: Leaf) => true
      case (x1: Fork, y1: Leaf) => x1.weight < y1.weight
      case (x1: Leaf, y1: Fork) => x1.weight < y1.weight
      case (x1: Fork, y1: Fork) => x1.weight < y1.weight
    }
  }

我尝试将 CodeTree 构造函数修改为:

I tried to modify CodeTree constructor to be:

  abstract class CodeTree(weight: Int)

这样我就可以直接比较 x 和 y，但是编译器说:

So that I can compare x and y directly, but compiler says:

构造函数 CodeTree 的参数不足:(weight: Int)patmat.Huffman.CodeTree"

还有其他方法可以缩短 sortCodeTreeFun 方法吗?

Is there another way to shorten sortCodeTreeFun method?

推荐答案

你可以简单地说:

def sortCodeTreeFun(x: CodeTree, y: CodeTree) = {
  (x, y) match {
    case (_: Leaf, _: Leaf)           => true
    case (x1: CodeTree, y1: CodeTree) => x1.weight < y1.weight
  }
}

并将抽象类CodeTree定义为

And define abstract class CodeTree as

abstract class CodeTree {
  def weight: Int
}

错误的原因是当你扩展一个带参数的类时，你需要提供参数.所以对于

The reason for the error is that when you extend an class that take a parameter you need to supply the parameter. So for

abstract class CodeTree(weight: Int)

您需要将其扩展为

case class Fork(left: CodeTree, right: CodeTree, chars: List[Char], weight: Int) extends CodeTree(weight)

这就是你得到的错误所说的:

That's what the error you were getting was saying:

"not enough arguments for constructor CodeTree: (weight: Int)"

这是因为您在扩展 CodeTree 时没有提供所需的参数 weight.

It was because you weren't supplying the required argument weight when extending CodeTree.

但是这种方法的问题是权重不是 CodeTree 的成员，因此不能从 CodeTree 类型的实例访问.也就是说，如果你这样做了:

The problem with this approach though is that weight is not a member of CodeTree and hence can not to be accessed from an instance of type CodeTree. That is to say, if you did:

 scala> Fork(...).asInstanceOf[CodeTree].weight
 <console>:11: error: value weight is not a member of CodeTree

因此，在您的模式匹配中，您将无法执行 x1.weight，因为 x1 的类型是 CodeTree 和 CodeTree 没有 weight.

So, in your pattern match, you wouldn't be able to do x1.weight since the type of x1 is a CodeTree and CodeTree does not have a weight.

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