Question

object Test extends Application {

  // compiles:
  Map[Int, Value](
    0 -> KnownType(classOf[Object]),
    1 -> UnknownValue())

  // does not compile:
  Map(
    0 -> KnownType(classOf[Object]),
    1 -> UnknownValue())
}
sealed trait Value {
  def getType: Option[Class[_]]
}
case class UnknownValue() extends Value {
  def getType = None
  // compiles if changed to:
  // def getType: Option[Class[_]] = None
}
case class KnownType(typ: Class[_]) extends Value {
  def getType = Some(typ)
}

The above code fails to compile. The compiler’s error message is:

Experiment.scala:10: error: type mismatch;
 found   : (Int, KnownType)
 required: (Int, Product with Value{def getType: Option[java.lang.Class[_$2]]}) where type _$2
    0 -> KnownType(classOf[Object]),
      ^
one error found

If I change the method declaration of UnknownValue to be def getType: Option[Class[_]] = None then also the Map() without type parameters compiles.

Why?

Answer 1

I posted the question to the scala-user mailing list and also they said that it’s a bug.