Path Dependency in Scala

Question

  trait B {
    type MyInnerType
    def foo: MyInnerType
  }      
  object B1 extends B {
    type MyInnerType = Double
    val foo = 3.0
  }

  trait A {
    type MyInnerType
    val b: B
    def foo(x: b.MyInnerType): MyInnerType
    def bar(y: MyInnerType): Unit
  }
  object A1 extends A {
    type MyInnerType = Int
    val b = B1
    def foo(x: b.MyInnerType) = 1
    def bar(y: MyInnerType) {}
  }
  object A2 extends A {
    type MyInnerType = String
    val b = B1
    def foo(x: b.MyInnerType) = "a"
    def bar(y: MyInnerType) {}
  }

  val as = Seq(A1, A2)
  as foreach { a => a.bar(a.foo(a.b.foo)) }   // wrong, a.foo(a.b.foo) infers to Any

However, if a.foo does not take parameters, everything works perfectly and a.foo infers to a.MyInnerType. It also works if I cast .asInstanceOf[a.MyInnerType]. Any explanations?

Answer 1

I'm running scala 2.9.1 and on the REPL I get this for as:

scala> val as = Seq(A1, A2)
as: Seq[ScalaObject with A{def foo(x: Double): Any; val b: B1.type; type MyInnerType >: java.lang.String with Int}] = List(A1$@6da13047, A2$@7168bd8b)

However, when I change it to val as:Seq[A] = Seq(A1, A2) I get:

scala> val as:Seq[A] = Seq(A1, A2)
as: Seq[A] = List(A1$@6da13047, A2$@7168bd8b)

scala> as foreach { a => a.bar(a.foo(a.b.foo)) }

Sometimes (all the time) scala has trouble inferring types, so you have to annotate what you want. I go to the REPL often to find out what's really going on.