Question

我有参数化类型的课程

abstract class Worker[T] {
  def conf1: ...
  def conf2: ...
  def doWork ...
}

abstract class SpecializedWorker[T: TypeTag] extends Worker[T] {
  //some behavior overriden (used fields from Trait that i want create)
}

我想创建可以混合到工人的特质。

trait Extension {
  self: Worker[_] =>      

  def someParameter: ... // only several workers can have that. thats why i need trait
  def produceSpecializedWorker = new SpecializedWorker[???]() {}
}

如何从self中提取类型信息以替换???

Answer 1

这是一种提取类型参数的方法：

trait Extension {
  self: Worker[_] =>      
  def mkSpWorker[T](implicit ev: this.type <:< Worker[T]) = new SpecializedWorker[T]() {}
}

但我不推荐:)你可以在Worker中定义一个可以在Extension中使用的类型成员。

abstract class Worker[T] {
  type TT = T
}

trait Extension {
  self: Worker[_] =>      
  def mkSpWorker = new SpecializedWorker[TT]() {}
}

或者你可以考虑给Extension一个类型参数。我认为这不是什么大问题。

在scala中从self中提取参数化类型

1 个答案: