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Home/ Questions/Q 9238817
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Editorial Team
  • 0
Editorial Team
Asked: June 18, 20262026-06-18T07:50:55+00:00 2026-06-18T07:50:55+00:00

I have been trying: class Foo(bar: Int) vs: class Foo(private val bar: Int) and

  • 0

I have been trying:

class Foo(bar: Int)

vs:

class Foo(private val bar: Int)

and they seem to behave the same although I couldn’t find anywhere saying that (bar:
Int)
expands to (private val bar: Int) so my question is, are these
identical/similar?

On a side note, I have been trying to use -Xprint:typer on these code pieces and they
produce the same code except for an extra line in the second one. How do I
read that extra line?

..
class Foo extends scala.AnyRef {
  <paramaccessor> private[this] val bar: Int = _;
  def <init>(bar: Int): this.Foo = {
    Foo.super.<init>();
    ()
  }
}
..


..
class Foo extends scala.AnyRef {
  <paramaccessor> private[this] val bar: Int = _;
  <stable> <accessor> <paramaccessor> private def bar: Int = Foo.this.bar;
  def <init>(bar: Int): this.Foo = {
    Foo.super.<init>();
    ()
  }
}
..
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  1. Editorial Team
    Editorial Team
    2026-06-18T07:50:56+00:00Added an answer on June 18, 2026 at 7:50 am

    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.

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