Scala generics type extends Comparable interface

Question

The following scala code works fine:

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))

But the next compiles with error:

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

Error: type arguments [B] do not conform to class C's type parameter bounds [T <: Comparable[T]]

Actually class B has also a compareTo member. How can I change class C's definition to make it compatible with class B? Thanks!

Mario Galic Mario Galic · Accepted Answer · 2019-06-10T20:58:51

Consider alternative implementation using Ordering type class like so

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)
}

which outputs

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

Scala generics type extends Comparable interface

2 Answers