如何使用特定类型创建List的包装器

一种可能的解决方案,感觉相当重量级,是：

import scala.collection._
import generic.{CanBuildFrom,SeqForwarder}
import mutable.{Builder,ListBuffer}

class MyList(list: List[Int]) extends immutable.LinearSeq[Int]
                                 with LinearSeqLike[Int,MyList]
                                 with SeqForwarder[Int] {
  override def newBuilder: Builder[Int,MyList] = MyList.newBuilder
  protected override def underlying = list
}

object MyList {
  def newBuilder: Builder[Int,MyList] =
    new ListBuffer[Int] mapResult(new MyList(_))

  implicit def canBuildFrom: CanBuildFrom[MyList,Int,MyList] = {
    new CanBuildFrom[MyList,MyList] {
      def apply(from: MyList) = from.newBuilder
      def apply() = newBuilder
    }
  }
}

val l1 = new MyList(List(1,2,3))

println(l1.isInstanceOf[MyList])
println(l1.map(_ + 1).isInstanceOf[MyList])
println(l1.filter(_ == 2).isInstanceOf[MyList])

是否有更好/更简单的方法来创建这样的包装器,还是我错过了MyList实现中的任何重要内容？

编辑：后续问题是：可以将整个包装逻辑放入ListWrapper类或特征中,以便上面的MyList可以像这样实现：

class MyList extends ListWrapper[Int,MyList]

object MyList extends ListWrapperFactory[Int,MyList]