任何对象的类型转换:模式匹配vs Try()

Any

让我们看一个使用模式匹配的示例:

abstract class A {
  def operation(input: Any): Any
}

class B extends A {
  // In this class, the input parameter is expected to be a Seq[Any]
  def operation(input: Any): Any = {
    input match {
      case _: Seq[Any] => Option(input.asInstanceOf[Seq[Any]]))
      case _ => None
    }
  }
}

class C extends A {
  // In this class, the input parameter is expected to be a Map[String, Any]
  def operation(input: Any): Any = {
    input match {
      case _: Map[String, Any] => Option(input.asInstanceOf[Map[String, Any]]))
      case _ => None
    }
  }
}

Try() 功能:

class B extends A {
  // In this class, the input parameter is expected to be a Seq[Any]
  def operation(input: Any): Any = {
    Try(input.asInstanceOf[Seq[Any]]).toOption
  }
}

class C extends A {
  // In this class, the input parameter is expected to be a Map[String, Any]
  def operation(input: Any): Any = {
    Try(input.asInstanceOf[Map[String, Any]]).toOption
  }
}

在Scala中,这些选项中哪一个是最佳实践,而且计算成本更低?有没有其他方法可以更有效地实现这个想法?