scala - Extending Enum in scalaz and implicit parameters with subtypes -

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• typeclasses , inheritance in scalaz 1 answer

i playing scalaz , thought extend enum type class make myself understand scalaz better. wrote this:

    sealed abstract trait counter     case object first extends counter     case object second extends counter     case object third extends counter      implicit val enumcounter: enum[counter] = new enum[counter] {       override def succ(a: counter): counter = match {         case first => second         case second => third         case third => first       }        override def pred(a: counter): counter = match {         case first => third         case second => first         case third => second       }        override def order(x: counter, y: counter): ordering = {         val map = map[counter, int](first -> 0, second -> 1, third -> 2)         implicitly[order[int]].order(map(x), map(y))       }     }      println(first |=> third)     println(first.succ) 

but turns out doesn't work hoped will. first not have succ nor |=>, because i've created enum[counter] not enum[first]. if write first.asinstanceof[counter].succ starts resolve. obvious me. how can implement enum typeclass in simple way? not want declare separate implicit values each of enum[first], enum[second]....

there 2 possible solutions thinking of:

1) make scala resolve enum[first] enum[counter]. cannot seem understand how can possible enum can nonvariant

2) maybe there solution in scalaz?

otherwise enum typeclass starts quite limited, not supports enum sounds weird.

i not sure how question belongs scalaz, depends on whether solution (1) or (2). if solution (1) - question pure scala.

i rethought question , rephrased lot - , received answer, please see this: typeclasses , inheritance in scalaz