scala - Defining an ordering on traits with abstract types which compiles only for compatible types? -

assume following trait:

trait {   type b        } 

is there way of making ordered type, a's same b's can compared, , enforced in compile time?

yes, via implicit (with type alias make things little more dry),

type aa[t] = { type b = t }  implicit def aisordered[t](a : aa[t]) = new ordered[aa[t]] {   def compare(that : aa[t]) = 0 } 

sample repl session,

scala> val ai1 = new { type b = int } ai1: java.lang.object a{type b = int} = $anon$1@1ec264c  scala> val ai2 = new { type b = int } ai2: java.lang.object a{type b = int} = $anon$1@1a8fb1b  scala> val ad = new { type b = double } ad: java.lang.object a{type b = double} = $anon$1@891a0  scala> ai1 < ai2 res2: boolean = false  scala> ai1 < ad <console>:16: error: type mismatch;  found   : ad.type (with underlying type java.lang.object a{type b = double})  required: aa[int]        ai1 < ad              ^ 

edit ...

thanks implicit definitions in scala.math.lowpriorityorderingimplicits defintion sufficient provide corresponding ordering type class instances. allows use types require orderings, eg. scala.collection.sortedset,

scala> implicitly[ordering[aa[int]]] res0: ordering[aa[int]] = scala.math.lowpriorityorderingimplicits$$anon$4@39cc63  scala> import scala.collection.sortedset import scala.collection.sortedset  scala> val s = sortedset(ai1, ai2) s: scala.collection.sortedset[java.lang.object a{type b = int}] = treeset($anon$1@1a8fb1b)