Scala 协方差分配给父类(super class)型

Question

我正在尝试找出 scala 协变和逆变。可能会混淆两个不同的概念。 处理下面的代码:

sealed trait Algorithm[ +T <: Model, P <: Model, R <: AnyVal] {
  def name: String
  def train(trainingData: DenseMatrix[Double]): T
  def predict(row: DenseVector[R], mlModel : P): R
}

然后我有两种算法类型声明为:

case class LibLinear() extends Algorithm[Active_Linear, Active_Linear, Double] {
  override val name = "libLinear"
  override def train(trainingData: DenseMatrix[Double]): Active_Linear = {
    ........
  }
  override def predict(row: DenseVector[Double], model: Active_Linear): Double = {
    ..........
  }
}


case class SVM() extends Algorithm[Volume_SVM, Volume_SVM, Double] {
  override val name = "libSVM"
  override def train(trainingData: DenseMatrix[Double]): Volume_SVM = {
    ..........
  }
  override def predict(row: DenseVector[Double], model: Volume_SVM): Double = {
    ...........
  }
}

其中Active_Linear和Volume_SVM都是Model的子类型。

现在我不能这样做:

val algorithm: Algorithm[Model, Model, Double] =  SVM()

SVM 是Algorithm 的子类型，Volume_SVM 是Model 的子类型。我们用协变和逆变符号声明算法。

Answer 1

这是因为算法仅在涉及 T 时才是协变的。 P 和 R 定义为不变式；您必须根据需要在每个代码前面添加 + 或 -，然后相应地修改代码。根据您的最终作业，我做了一些假设，这就是我得出的结论:

sealed trait Algorithm[ +T <: Model, +P <: Model, +R <: AnyVal] {
  def name: String
  def train(trainingData: DenseMatrix[Double]): T
  def predict[U >: R, V >: P](row: DenseVector[U], mlModel : V): U
}

case class LibLinear() extends Algorithm[Active_Linear, Active_Linear, Double] {
  override val name = "libLinear"
  override def train(trainingData: DenseMatrix[Double]): Active_Linear = {
    ...
  }

  override def predict[U >: Double, V >: Active_Linear](row: DenseVector[U], model: V): U = {
    ...
  }
}


case class SVM() extends Algorithm[Volume_SVM, Volume_SVM, Double] {
  override val name = "libSVM"
  override def train(trainingData: DenseMatrix[Double]): Volume_SVM = {
      ...
  }
  override def predict[U >: Double, V >: Volume_SVM](row: DenseVector[U], model: V): U = {
      ...
  }
}

你的最后一个作业就可以正常工作了。