How to initialize the value from trait in subtype?

Question

If I write :

trait T {
  val t = 3
  val u = 1::t::Nil
}

class U extends T {
  override val t = 2
}

(new U).u

it shows this.

List(1, 0)

How should I change the above code to make it display the following:

List(1, 2)

i.e. override val t sets the value of t for u in the trait T?

Answer 1

Try using an early initializer:

scala> trait T {
     |   val t = 3
     |   val u = 1::t::Nil
     | }
defined trait T

scala> class U extends {
     | override val t = 2;
     | } with T
defined class U

scala> (new U).u
res1: List[Int] = List(1, 2)

See e.g. here for more information about early initialization.

Answer 2

All scala declarative style is just an illusion. Scala is build upon a jvm and works like java.

Evetything is a class and should be independent on its usage (java is not c++ and supports incremental build with its pros and cons). Every trait has its own initialization code and multi-trait class runs respective initialization code one by one. If you use some AnyRef that is declared only in a subclass than that its value will be set for null during initialization.

I guard myself with specifing convention rule: every val should be either final or lazy (why using plain val in non-final classes) . So I don't care about initialization order and may pretend further that I'm using declarative language.

Also I'm using option -Xcheckinit: Add runtime check to field accessors.

Answer 3

One way to do this is to delay evaluation of u by using def or lazy val as follows:

trait T {
  def t = 3
  def u = 1::t::Nil
}

class U extends T {
  override def t = 2
}

(new U).u

or

trait T {
  val t = 3
  lazy val u = 1::t::Nil
}

class U extends T {
  override val t = 2
}

(new U).u 

The differences are as follows:

• val makes an expression evaluate during initialization
• def makes an expression evaluate each time u is used
• lazy val makes it evaluated on first u usage and caches the result