Scala Ordering[T] and Serializable

Question

Given a case class like this:

case class KVPair[
    K <: Ordering[K] with Serializable, 
    VV <: Serializable
](key: K, value: VV)

How would I go about constructing a KVPair with Longs, something like this:

def createKVPair(): KVPair[Long, Long] = {
    KVPair(1L, 1L)
}

The compiler keeps complaining with:

Long does not conform to method apply's type parameter bounds.

I assume this is because `Long` is not an `Ordering`. Have you tried `K <% Comparable`? — Gregor Raýman, Apr 30 '15 at 12:41
I believe B. Kemmer is right. Long is per definition Ordering, and Ordering inherits from Serializable. It is beyond me why this won't work. — Pieter-Jan, Apr 30 '15 at 12:54

score 3 · Accepted Answer · edited May 23 '17 at 12:21

There are 2 things going on here. First of all Long does not extend Ordering[Long]. The page you linked from the docs in the comments is the documentation of an implicit object called Long in the scala.math.Ordering package.

However, that object is very relevant here. Since that object is in implicit scope, you can use a context bound to ensure that there is an instance of Ordered[Long] is available, which will allow code to order your Longs.

case class Key[K : Ordering](key: K)

This will allow you to do Key(1L).

Your other problem is that Long is also not Serializable, so the second generic parameter is also problematic. Long is implicitly viewable as a java.io.Serializable, which might be of interest to you here. You can use a view bound to ensure that K and V have this constraint:

type JS = java.io.Serializable
case class KVPair[K <% JS : Ordering, V <% JS](key: K, value: V)

The context-bound and view-bound syntax make this equivalent to:

 case class KVPair[K, V](key: K, value: V)(
    implicit kev1: Ordering[K], 
             kev2: K => JS, 
             vev1: V => JS
 )

You can read more about context bounds and view bounds on this answer.

Michael Zajac · Answer 2 · 2015-04-30T13:45:33.643

scala.Long is not an Ordering. The type bound K <: Ordering[K] means K should must be bounded by the Ordering type. And scala.Long most certainly does not extend the type class Ordering. There exists an instance of an Ordering[Long], but scala.Long does not extend it.

What you want is evidence of an Ordering[K], ie., a context bound. Something like this:

case class KVPair[K : Ordering , VV](key: K, value: VV)

def createKVPair(): KVPair[Long, Long] = {
    KVPair(1L, 1L)
}

The above case class is syntactic sugar for:

case class KVPair[K, VV](key: K, value: VV)(implicit ev: Ordering[K])

You may have noticed that I've also removed the Serializable bound. This is because Long is not Serializable, so it cannot possibly work with that constraint.

scala> implicitly[Long <:< Serializable]
<console>:26: error: Cannot prove that Long <:< Serializable.
              implicitly[Long <:< Serializable]

Scala Ordering[T] and Serializable

2 Answers2