Scala: var to val - redesign

Question

In my code base, I want to refractor away from vars. The code base structure follows the format:

class Animal {
    var name : Option[String] = None
    var owner : Option[String] = None
}

case class Dog(breed: String) extends Animal {
    //Dog logic
}

The main reason for this design is that information is not available all at the same time.

So, from service A, which deserializes a json, I receive an Animal (either Animal, Dog)

val animal = new Animal 
animal.name = "My little animal"

and then

def update(animal: Animal) : Unit {
    animal match {
        case a : Animal => a.owner = Some("Joe") // call to service B
        case _          => //handle dog
    }
}

update(animal)

The question is: How can I redesign it to avoid mutable state in Animal?

Write a copy method in Animal? There are a bunch of fields, which might generate some boilerplate
Composition?

Making the two classes case?

case class Animal (...)
case class Dog(animal: Animal, breed: String)

Edit

Animal as a trait

I still need a concrete implementation of Animal, as I will have Dogs and Animals (which are really Animal and no subtype)

Problems with copy

The built in case class copy method, does NOT update the values of the Animal class. So defaults values will be used - not what I want. - yes, it is possible to create a case class with all the fields, but is it practical if we have 100+ fields?

Case classes include a copy method https://stackoverflow.com/questions/7249396/how-to-clone-a-case-class-instance-and-change-just-one-field-in-scala — ameer, Dec 21 '17 at 18:40

joesan · Answer 1 · 2017-12-21T18:56:05.803

0

You could design this as an ADT (Abstract Data Type) like this:

trait Animal {
  def name: Option[String]
  def owner: Option[String]
}

case class Dog(dog: Option[String] = None, owner: Option[String] = None) extends Animal

With a case class you have the copy method by default!

edited Dec 21 '17 at 18:56

answered Dec 21 '17 at 18:46

joesan

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If `Animal` is a `trait` the deserialization fails to create an `Animal` – Bruno Dec 21 '17 at 18:52
I edited by adding the extends clause to Dog case class – joesan Dec 21 '17 at 18:56

score 0 · Answer 2 · answered Dec 22 '17 at 00:26

A good solution could be to make a trait which specifies Animal behavior,

sealed trait Animal {
  val name: Option[String]
  val owner: Option[String]
}

and then make case classes to provide type constructors for your animal instances.

case class Dog(name: Option[String], owner: Option[String]) extends Animal

Now, update (changeOwner) will take an Animal and return another Animal likewise:

def changeOwner(animal: Animal): Animal = animal match {
  case Dog(name, owner) => Dog(name, Some("Joe"))
  case _  => animal
}

And you will use it as follows:

val dog: Animal = Dog(Some("Doggie"), Some("Jack"))
val newDog: Animal = changeOwner(dog)

Also, from your code:

case a : Animal => a.owner = Some("Joe") // call to service B

Be careful with this, as the first case of your match will take all animals and thus the other cases, for example, your Dog case, will be unreachable.

I edited explaining why `trait Animal` does not suit me and the drawbacks of `copy` — Bruno, Dec 22 '17 at 08:35

Scala: var to val - redesign

2 Answers2