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Scala 如何将case类与varargs匹配?_Scala - Fatal编程技术网

Scala 如何将case类与varargs匹配?

Scala 如何将case类与varargs匹配?,scala,Scala,我有以下Akka演员代码: class GenerateBoardActor extends Actor { import GenerateBoardActor._ def receive = { case GenerateBoard(configuration: Seq[Configuration.PiecesConfigSpec]) => { sender ! Board(configuration: _*) } case generat

我有以下Akka演员代码:

class GenerateBoardActor extends Actor {
  import GenerateBoardActor._

  def receive = {
    case GenerateBoard(configuration: Seq[Configuration.PiecesConfigSpec]) => {
      sender ! Board(configuration: _*)
    }

    case generateBoard: GenerateBoard => {
      sender ! Board(generateBoard.configuration: _*)
    }
  }
}

object GenerateBoardActor {
  case class GenerateBoard(configuration: Configuration.PiecesConfigSpec*)
}
我认为这两个case子句应该是等价的,但只有第二个是匹配的。是否可以使用第一个子句的语法将case类与varargs匹配?或者第一个子句的正确语法是什么?

正确的语法是:

case GenerateBoard(configuration @ _*) =>