Do scala constructor parameters default to private val?

Question

bar: Int

This is barely a constructor parameter. If this variable is not used anywhere except the constructor, it remains there. No field is generated. Otherwise private val bar field is created and value of bar parameter is assigned to it. No getter is created.

private val bar: Int

Such declaration of parameter will create private val bar field with private getter. This behavior is the same as above no matter if the parameter was used beside the constructor (e.g. in toString() or not).

val bar: Int

Same as above but Scala-like getter is public

bar: Int in case classes

When case classes are involved, by default each parameter has val modifier.

In the first case, bar is only a constructor parameter. Since the main constructor is the content of the class itself, it is accessible in it, but only from this very instance. So it is almost equivalent to:

class Foo(private[this] val bar:Int)

On the other hand, in the second case bar is a normal private field, so it is accessible to this instance and other instances of Foo. For example, this compiles fine:

class Foo(private val bar: Int) {
  def otherBar(f: Foo) {
    println(f.bar) // access bar of another foo
  }
}

And runs:

scala> val a = new Foo(1)
a: Foo = Foo@7a99d0af

scala> a.otherBar(new Foo(3))
3

But this doesn't:

class Foo(bar: Int) {
  def otherBar(f: Foo) {
    println(f.bar) // error! cannot access bar of another foo
  }
}