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scala - Scala 泛型类型扩展了 Comparable 接口

问题描述

以下 scala 代码可以正常工作：

class A(val value : Int) extends Comparable[A]
{
    override def compareTo(o: A): Int = this.value.compareTo(o.value)
}

class B(value : Int) extends A(value)

class C[T <: Comparable[T]](val value : T) extends Comparable[C[T]]
{
    override def compareTo(o: C[T]): Int = value.compareTo(o.value)
}

val a1 = new A(1)
val a2 = new A(2)
println(a1.compareTo(a2))

val b1 = new B(1)
val b2 = new B(2)
println(b1.compareTo(b2))

val ac1 = new C[A](a1)
val ac2 = new C[A](a2)
println(ac1.compareTo(ac2))

但下一个编译错误：

val bc1 = new C[B](b1)
val bc2 = new C[B](b2)
println(bc1.compareTo(bc2))

错误：类型参数 [B] 不符合类 C 的类型参数界限 [T <: Comparable[T]]

实际上 B 类也有一个 compareTo 成员。如何更改 C 类的定义以使其与 B 类兼容？谢谢！

标签： scalagenericsf-bounded-polymorphism

解决方案

Ordering考虑使用像这样的类型类的替代实现

class A(val value: Int)
class B(value: Int) extends A(value)

class C[T](val value: T)

implicit val aOrdering = new Ordering[A] {
  def compare(a1: A, a2: A): Int = a1.value compare a2.value
}

implicit val bOrdering = new Ordering[B] {
  def compare(b1: B, b2: B): Int = b1.value compare b2.value
}

implicit class CompareToA[T <: A](`this`: T) {
  def compareTo(that: T)(implicit o: Ordering[T]): Int = o.compare(`this`, that)
}

implicit class CompareToC[T](`this`: C[T]) {
  def compareTo(that: C[T])(implicit o: Ordering[T]): Int = o.compare(`this`.value, that.value)
}

哪个输出

val b1 = new B(1)
val b2 = new B(2)
val bc1 = new C[B](b1)
val bc2 = new C[B](b2)
bc1.compareTo(bc2) // res3: Int = -1

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