spec should explicitly state that qualified private members can be inherited

[scabug]

For the following code:

object p {
  trait A {
    private[p] val m = 1
  }
}

trait B {
  val m = "1"
}
trait Mix extends p.A with B

I get the message:

  value m in trait A of type Int  and
  value m in trait B of type String
 (Note: this can be resolved by declaring an override in trait Mix.)
 trait Mix extends p.A with B

Maybe my expectations are incorrect but I expected the package protected member to never conflict with another member outside of the definition package.

[scabug]

Imported From: https://issues.scala-lang.org/browse/SI-9321?orig=1
Reporter: @etorreborre
Affected Versions: 2.11.6

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@retronym said:
Here's a tighter look at this. Relevant spec extracts inline:

object p {
  trait A {
    private val privateMember = 1
    private[p] val qualifiedPrivateMember = 1
  }

  // The private modifier can be used with any definition or declaration in a template.
  // Such members can be accessed only from within the directly enclosing template and
  // its companion module or companion class. They are not inherited by subclasses and
  // they may not override definitions in parent classes.
  def usePrivate(b: Mix) =
    b.privateMember // not a member

  // The modifier can be qualified with an identifier C (e.g. private[C]) that must
  // denote a class or package enclosing the definition. Members labeled with such a
  // modifier are accessible respectively only from code inside the package C or only
  // from code inside the class C and its companion module.

  def useQualifiedPrivate(b: Mix) =
    b.qualifiedPrivateMember // allowed, but should it be?
}

trait Mix extends p.A

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@retronym said:
We're clarifying the spec to say this is as designed: scala/scala#4512

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@etorreborre said:
Thanks Jason for looking into this.