在 Scala 中匹配多个案例类 [英] Match multiple cases classes in scala

问题描述

我正在对某些案例类进行匹配，并希望以相同的方式处理其中的两个案例.像这样:

I'm doing matching against some case classes and would like to handle two of the cases in the same way. Something like this:

abstract class Foo
case class A extends Foo
case class B(s:String) extends Foo
case class C(s:String) extends Foo


def matcher(l: Foo): String = {
  l match {
    case A() => "A"
    case B(sb) | C(sc) => "B"
    case _ => "default"
  }
}

但是当我这样做时，我得到了错误:

But when I do this I get the error:

(fragment of test.scala):10: error: illegal variable in pattern alternative
    case B(sb) | C(sc) => "B"

我可以让它工作，因为我从 B 和 C 的定义中删除了参数，但是我如何与参数匹配?

I can get it working of I remove the parameters from the definition of B and C but how can I match with the params?

推荐答案

看来你不关心 String 参数的值，想把 B 和 C 一视同仁，所以:

Looks like you don't care about the values of the String parameters, and want to treat B and C the same, so:

def matcher(l: Foo): String = {
  l match {
    case A() => "A"
    case B(_) | C(_) => "B"
    case _ => "default"
  }
}

如果你必须，必须，必须提取参数并在同一个代码块中处理它们，你可以:

If you must, must, must extract the parameter and treat them in the same code block, you could:

def matcher(l: Foo): String = {
  l match {
    case A() => "A"
    case bOrC @ (B(_) | C(_)) => {
      val s = bOrC.asInstanceOf[{def s: String}].s // ugly, ugly
      "B(" + s + ")"
    }
    case _ => "default"
  }
}

虽然我觉得把它分解成一个方法会更清晰:

Though I feel it would be much cleaner to factor that out into a method:

def doB(s: String) = { "B(" + s + ")" }

def matcher(l: Foo): String = {
  l match {
    case A() => "A"
    case B(s) => doB(s)
    case C(s) => doB(s)
    case _ => "default"
  }
}

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