Scala：A 扩展 Ordered[A]

Question

我刚刚开始学习 Scala，为了练习，我决定创建一个 Pair[A, B] 类，该类在排序时首先按 As 排序，然后按 Bs 排序。我的第一次尝试是这样的：

case class Pair[A <: Ordered[A], B <: Ordered[B]](val left: A, val right: B) extends Ordered[Pair[A, B]]
{
  override def compare(that: Pair[A, B]) = {
      val leftCompare = this.left.compare(that.left)
      if (leftCompare == 0)
        this.right.compare(that.right)
      else
        leftCompare
    }
}

object Main extends App
{
  List(Pair(1, "a"), Pair(5, "b"), Pair(5, "a"), Pair(1, "b")).sorted
}

这就是我想要的方式，真的。我想扩展 Ordered 并让它工作，所以我可以做像 List(whatever).sortWith(_ < _) 或 List(whatever).sorted 之类的事情，就像我在 Main 中写的那样。我收到以下错误：

pair.scala:14: error: inferred type arguments [Int,String] do not conform to method apply's type parameter bounds [A <: Ordered[A],B <: Ordered[B]]
  List(Pair(1, "a"), Pair(5, "b"), Pair(5, "a"), Pair(1, "b")).sorted

这对于列表中的每一对：

pair.scala:14: error: type mismatch;
 found   : Int(1)
 required: A
  List(Pair(1, "a"), Pair(5, "b"), Pair(5, "a"), Pair(1, "b")).sorted

这也是，我不了解：

pair.scala:14: error: diverging implicit expansion for type scala.math.Ordering[Pair[_ >: A with A with A with A <: scala.math.Ordered[_ >: A with A with A with A <: scala.math.Ordered[_ >: A with A with A with A]], _ >: B with B with B with B <: scala.math.Ordered[_ >: B with B with B with B <: scala.math.Ordered[_ >: B with B with B with B]]]]
starting with method $conforms in object Predef
  List(Pair(1, "a"), Pair(5, "b"), Pair(5, "a"), Pair(1, "b")).sorted

我设法让它排序，但只能通过编写知道它们正在排序的类型的排序函数，而不是仅仅知道它们正在排序可排序的类型。正如我所提到的，这就是这些“了解类型”的版本：

case class Pair[A, B](val left: A, val right: B)
{}

object Main extends App
{
  val pairs = Array(Pair(1, "a"), Pair(5, "b"), Pair(5, "a"), Pair(1, "b"))
  Sorting.quickSort(pairs)(Ordering[(Int, String)].on(x => (x.left, x.right)))
  println(pairs.toList)
}

和

case class Pair[A, B](val left: A, val right: B)
{}

object Main extends App
{
  val intStringSort = (x: Pair[Int, String], y: Pair[Int, String]) => {
    val intCompare = x.left - y.left
    if (intCompare == 0)
      x.right.compare(y.right) < 0
    else
      intCompare < 0
  }
  println(List(Pair(1, "a"), Pair(5, "b"), Pair(5, "a"), Pair(1, "b")).sortWith(intStringSort))
}

提前致谢。

Answer 1

使用<%代替<:

case class Pair[A <% Ordered[A], B <% Ordered[B]]

因为既不是Int也不String是Ordered。

更新：

但是存在从Intto RichIntwhich isOrdered[Int]和 from Stringto StringOpswhich is 的隐式转换Ordered[String]。

<%意味着对象可以隐式转换为定义的类型。例如，A <% Ordered[A]对于Int装置有从一个隐式转换Int到Ordered[Int]它是：

implicit def intWrapper(x: Int)         = new runtime.RichInt(x)

在scala.Predef该自动导入。RichInt是Ordered[Int]。对 执行类似的步骤String。