CODEWITHSUNDEEP
scalainheritanceconstructorpolymorphismsuperclass
user826955
user826955

Reputation: 3206

Using alternative super class constructor in child class instantiation

I have a base class with two constructors, and a child class which has one constructor. Is it possible to instantiate a child class using the second base class constructor?

Example code:

abstract class RuleCondition(rule:Rule, field:String, equal:Boolean, inverted:Boolean)
{
  // alternate constructor with RuleValue instead of static comparation value

  def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = ???
 }

class RuleConditionAbove(rule:Rule, field:String, comparationValue:Long, equal:Boolean = false, inverted:Boolean = false)
  extends RuleCondition(rule, field, equal, inverted)
{
    // ...
}

Now I can do this:

val myAboveCondition = new RuleConditionAbove(rule, "bla", 10, true, false)

but I cannot do this:

val myAboveCondition = new RuleConditionAbove(rule, "bla", RuleValue(...), true, false)

because the alternative constructor of RuleCondition base class is not visible. It will be visible once I add this to the child class:

def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = this(rule, field, ref, equal, inverted)

Would this be the only/usual way of solving this issue, or is there something smarter which involves less copy & past code? (since I have a ton of child classes of the same pattern)

[edit] To clarify, the second constructor would be the same in every child class, thus I would like to have it implemented only once in the base class. However still having to put another constructor in each child class would defeat this purpose somehow, and thus I would not have two constructors in the base class but rather only in all child classes.

Upvotes: 1

Views: 250

Answers (2)

Ivan Stanislavciuc
Ivan Stanislavciuc

Reputation: 7275

You have to add a constructor definition in every child class as you described. 

def this(rule:Rule, field:String, ref:RuleValue, equal:Boolean = false, inverted:Boolean = false) = this(rule, field, ref, equal, inverted)

Imagine that a child class defines new fields that are not available in the base class. Creating child class with a base constructor would not define such fields and leave the instance of the class partially initialised.

If your base constructor has valuable logic, it makes sense to keep it in the base class. And just "link" it to the base constructor in the child class.

Upvotes: 0

Tim
Tim

Reputation: 27356

Is it possible to instanciate a child class using the [second] base class constructor?

No.

You can never use a superclass constructor to create an instance of a subclass. You have to call a constructor for the class that you are creating. The subclass constructor must call a constructor for the superclass, but you can't call it directly.

So the reason that you can do this

val myAboveCondition = new RuleConditionAbove(rule, "bla", 10, true, false)

is that RuleConditionAbove has a constructor with those arguments. It has nothing to do with the fact that RuleCondition has a constructor with the same arguments.

And the reason that you can't do this

val myAboveCondition = new RuleConditionAbove(rule, "bla", RuleValue(...), true, false)

is that RuleConditionAbove does not have a constructor with those arguments.

Upvotes: 4

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