Scala traits mixin order and super call

问题

I have this code:

trait base{
  def msg: Unit= {
    println{"base"}
  }
}

trait foo extends base {
  abstract override def msg: Unit ={
    super.msg
    println("foo")
  }
}

class base2{
  def msg:Unit = {
    println{"base 2"}
  }
}

class test extends base2 with foo{
   override def msg: Unit ={
    super.msg
    println("done")
  }
}

If I call (new test).msg, this prints out things like: base, foo, done

However, if I change the base trait to:

trait base{
  def msg: Unit
}

it prints out things like: base 2, foo, done

I understand the order of with is from right to left (last one comes first) but how about extends? How come sometimes it prints base2, but sometimes base?

回答1:

When you omit the implementation, base is a template of a trait and has different evaluation rules. See the Scala specification

回答2:

Scala has something called type linearization. It defines initialization order. Read here http://eed3si9n.com/constraining-class-linearization-in-Scala

来源：https://stackoverflow.com/questions/27569901/scala-traits-mixin-order-and-super-call