ExprType paths must not be considered stable

[samuelgruetter]

dotty considers that p of type => P is stable, so that p.T is a valid type, but that's unsound:

object test {
  trait Consumer {
    type T
    def eat(t: T): Unit
    val defaultValue: T
  }
  class IntConsumer extends Consumer {
    type T = Int
    def eat(i: Int): Unit = println(i*i)
    val defaultValue = 0
  }
  class StringConsumer extends Consumer {
    type T = String
    def eat(s: String): Unit = println(s.length)
    val defaultValue = ""
  }

  var counter: Int = 0
  def unstableConsumer: Consumer = {
    counter += 1
    if (counter % 2 == 0) new IntConsumer else new StringConsumer
  }

  def feed(c: => Consumer): Unit = {
    val v: c.T = c.defaultValue // v: String = ""
    c.eat(v) // c: IntConsumer, will crash
  }

  feed(unstableConsumer)
}

Scala 2 rejects it:

error: stable identifier required, but c found.
    val v: c.T = c.defaultValue // v: String = ""
           ^

but dotty accepts it.

[odersky]

Fixed by way of #1051.