Overriding val in scala

Question

I'm confused by the following:

class A(val s: String) {
 def supershow {
   println(s)
 } 
}


class B(override val s: String) extends A("why don't I see this?"){
  def show {
    println(s)
  }
  def showSuper {
    super.supershow
  }
}

object A extends App {
  val b = new B("mystring")
  b.show
  b.showSuper
}

I was expecting:

mystring
why don't I see this?

But I get:

mystring
mystring

In java if you override, or 'shadow' a variable in a super class, the super class has its own variables. But here, even though I think I'm explicitly initializing the parent with a different string, the parent gets set to the same value as the subclass?

Answer 1

In scala val is similar to getter method in java. You can even override def with val.

If you need something similar to field from java you should use private[this] val:

class A(private[this] val s: String) {
  def superShow() = println(s)
}
class B(private[this] val s: String) extends A("why don't I see this?") {
  def show() = println(s)
}

val b = new B("message")
b.show
// message
b.superShow()
// why don't I see this?