Parameterized method with Ordering?

Question

Now I am confused. I am quite new on Scala, having worked with it for a few weeks, I think I am getting familiar with it, but I am stuck on the apparently trivial following case.

I cannot find the Scala equivalent to this Java declaration :

public static <T extends Comparable<T>> List<T> myMethod(List<T> values) {
  // ...
  final List<T> sorted = new ArrayList<T>(values);
  Collections.sort(sorted);
  // ...
}

I thought the following would do :

def myMethod[A >: Ordering[A]](values: Seq[A]): Seq[A] = {
  // ...
  val sorted = values.sorted
  //
}

However, I get the following errors:

error: illegal cyclic reference involving type A

error: diverging implicit expansion for type scala.math.Ordering[A] starting with method Tuple9 in object Ordering

Where am I wrong?

it should be A **<:** Ordering[A], A is a subtype of (i.e. extends) Ordering[A]. That way it should compile. — Paolo Falabella, May 31 '11 at 13:28

score 7 · Answer 1 · edited May 23 '17 at 12:04

First of all, Ordering is akin to Comparator, not Comparable. The Scala equivalent to Comparable is Ordered. Next, extends equivalent is <:, not >:. The latter is equivalent to super -- T super COmparable<T>, which is not what you want. So, with both these fixes in, your code should look like this:

def myMethod[A <: Ordered[A]](values: Seq[A]): Seq[A] = {
  // ...
  val sorted = values.sorted
  //
}

However, this won't work for Seq[Int], for instance, since Int does not extend Ordered -- just like Java's int does not extend Comparable (or anything else, since it is not a class).

Scala has a work-around for that -- an implicit conversion from certain classes to an Ordered class. To use that, however, you need to use a view bound, which would make the code look like this:

def myMethod[A <% Ordered[A]](values: Seq[A]): Seq[A] = {
  // ...
  val sorted = values.sorted
  //
}

See <% instead of <:? That's the view bound.

Now, the current preference in Scala circles is to use context bounds instead of view bounds, as they are more flexible. That would mean using Ordering, in the manner described by the other answers.

score 5 · Accepted Answer · answered May 31 '11 at 13:49

5

That should be a context bound, as shown below.

scala> def myMethod[A : Ordering](values: Seq[A]): Seq[A] = values.sorted
myMethod: [A](values: Seq[A])(implicit evidence$1: Ordering[A])Seq[A]

answered May 31 '11 at 13:49

missingfaktor

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Parameterized method with Ordering?

2 Answers2