Question

Considering:

abstract class Base {
  def something() = println("Base")
}

trait TraitA extends Base {
    abstract override def something() = { super.something(); println("TraitA"); }
}

class Child extends Base {
    override def something() {
        println("Child")
    }
}

And then:

val x = new Child with TraitA
x.something()

I get:

Child
TraitA

But if I use:

class Child extends Base with TraitA {
    override def something() {
        println("Child")
    }
}

val x = new Child
x.something()

I’d get only:

Child

So what are the differentes of using a trait at construction site or declaration site? Can I have the first behavior (stackable traits) while extending/with’ing the trait in the declaration of the class?

Answer 1

Stackable traits in Scala refers to being able to mix in multiple
traits that work together to apply multiple modifications to a method.
This involves invoking super.theMethod and modifying its input and/or
output. But what is super in the context of a trait? The class (or
trait) the trait extends from? The class the trait is being mixed
into? It depends! All the mixed in traits and all the superclasses are
linearized. super invokes the nearest preceding definition further up
the chain.

read full post

In your case:
1. super.something() == Child.something()
2. super.something() == Base.something() and Child.something() overrides whole chain

Type of mixin doesn’t matter:

class Child2 extends Child with TraitA

val x2 = new Child2
x2.something() 
/* 
   Child
   TraitA
*/

class Child3 extends Base with TraitA
val x3 = new Child3
x3.something()
/* 
   Base
   TraitA
*/