Scala - val parameters

Question

Let's say I have an

object Error{
    abstract class Reason{def p:P}
    case class A(p:P) extends Reason
    case class B(p:P) extends Reason
    ...
}

And that now I want to add a special case of B:

object Error{
    abstract class Reason{def p:P}
    case class A(p:P) extends Reason
    class B(val p:P) extends Reason{
        def apply(p:P) = new B(p)
        def unappply(b:B) = 
            if(b == null) None
            else Some(b.p)
    }

    case class SpecialB(override val p:P) extends B(p)
    ...
}

In this change, why must I declare parameter p in B as val and why didn't I have to when B was a case class?

score 2 · Answer 1 · answered Jun 03 '15 at 07:32

2

All parameters of a case class constructor are exposed as public values, which is not the case for a plain class, for which val must be used to make the parameter public.

answered Jun 03 '15 at 07:32

cchantep

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score 1 · Answer 2 · edited May 23 '17 at 11:43

1

In a case class when you provide constructor parameter p, a field with public getter method with same signature as def p:P is created.

In a regular class you have to specify that parameter should have public methods via val or var. Otherwise it can be discarded after initialization. See Scala Constructor Parameters

edited May 23 '17 at 11:43

Community

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answered Jun 03 '15 at 07:37

Tyth

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Scala - val parameters

2 Answers2