scala - Define a type constraint with abstract type -

i try define type constraint abstract type. unfortunately, doesn't compile.

  sealed trait matchablevalue {     type     def value:      def assingleitemvalue : itemvalue   }    sealed trait itemvalue {     type     def value:   }    case class stringvalue(value: string) extends itemvalue {type = string}   case class stringmatchablevalue(value: string) extends matchablevalue{     type = string     override def assingleitemvalue =  stringvalue(value)   } 

unfortunately, 1 doesn't work

def assingleitemvalue[b <: itemvalue](implicit ev: =:= b#a) : b 

the aim of type constraint warned @ compile-time of such error :

  case class intvalue(value: int) extends itemvalue {type = int}   case class intmatchablevalue(value: int) extends matchablevalue{     type = int     def assingleitemvalue = stringvalue("error")   } 

you can accomplish type refinement (note method's return type):

sealed trait matchablevalue { self =>   type   def value:    def assingleitemvalue: itemvalue { type = self.a } }  sealed trait itemvalue {   type   def value: }  case class stringvalue(value: string) extends itemvalue { type = string } case class intvalue(value: int) extends itemvalue { type = int } 

now compiles:

case class stringmatchablevalue(value: string) extends matchablevalue {   type = string   def assingleitemvalue = stringvalue(value) } 

but doesn't:

case class stringmatchablevalue(value: string) extends matchablevalue {   type = string   def assingleitemvalue = intvalue(1) } 

which believe want.

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it's worth noting following common pattern when dealing type refinements:

sealed trait matchablevalue { self =>   type   def value:    def assingleitemvalue: itemvalue.aux[a] }  sealed trait itemvalue {   type   def value: }  object itemvalue {   type aux[a0] = itemvalue { type = a0 } } 

this same thing, syntax nice alternative if find needing write out type refinement lot.