Scala: "override protected val" results in error when defining case class constructor

Question

I am experiencing some strange thing related with case classes and override protected val.

Let's say trait T have a protected def (or val).

trait T {
  protected def s: String
  def print(): Unit = println(s)
}

I can override this method def s with val s. There's no problem.

class A(override val s: String) extends T

OK. It compiles. But this makes s public since there is no protected keyword.

class B extends T {
  override protected val s: String = "Show me the money"
}

OK. It compiles. (s of class B is protected)

class C(override protected val s: String) extends T

OK. It compiles. (s of class C is protected) Let's do these with case class.

case class D(override val s: String) extends T

OK. It compiles. (although s of class D is public)

case class E() extends T {
  override protected val s: String = "Show me the money"
}

OK. It compiles. (s of class E is protected)

case class F(protected val s: String) extends T

OK. It compiles. (s of class F is protected, and also s is overriden as it was not concrete in trait T. It was abstract, so override keyword is not necessary.)

case class G(override protected val s: String) extends T

Here, compile error appears. error: value s$1 overrides nothing

Basically, class F and class G are the same. class F overrides s and s is protected. This can be shown by that F("Something").print() will indeed print Something. override protected val can be used in case class as class E shows. It did not make errors in defining normal class C.

However, only defining class G results in a compile error.

Why does this thing happen?

I'm using Scala 2.10.3

Answer 1

This bug will be fixed in Scala 2.11.0-M8

https://issues.scala-lang.org/browse/SI-8132