How to implement Ordered properly - scala

Question

I have a trait
trait Weight {
def getWeight: Int
}
Multiple classes inherits it, example:
case class Test(n: Int) extends Weight {
override def getWeight: Int = n
}
Now i want to add sorting ability to all Weight subclasses. I added Ordered to Weight:
trait Weight extends Ordered[Weight] {
def getWeight: Int
override def compare(that: Weight): Int = this.getWeight.compareTo(that.getWeight)
}
Try sorting:
val seq = Seq(Test(1), Test(4), Test(3), Test(2))
seq.sorted // error
And it's not compiles:
Error:(74, 6) diverging implicit expansion for type
scala.math.Ordering[A$A254.this.Test] starting with method $conforms
in object Predef seq.sorted;}
^
Whats i am doing wrong?

Answer 1

Another solution a bit different than mdm. Since sorted takes an implicit of Ordering, you can do the following:
seq.sorted(Ordering[Weight])

Answer 2

Your solution does not work because Ordered[T] is invariant in T, meaning that Ordered[Weight] has no relationship with Ordered[A]. You would need to specify that in the sub-classes.
You could use an implicit Ordering rather than an Ordered.
trait Weight{
def getWeight : Int
}
object Weight{
implicit def ordering[T <: Weight] : Ordering[T] = Ordering.by(w => w.getWeight)
}
case class A(w : Int) extends Weight{
def getWeight = w
}
case class B(w : Int) extends Weight{
def getWeight = w
}
import Weight._
Seq(A(1),B(2),B(0),A(3),A(-3)).sorted
Will result in:
List(A(-3), B(0), A(1), B(2), A(3))
Note that this solution relies on an Ordering[Int] to be available (which is, by default).