Making traversal generic in scala - scala

Question

I know there are already some ways of making traversal generic in scala, but I'm trying to define a simple trait that encapsulates a traversable type. However, I'm having trouble getting the type signatures to work out. A first attempt would be something like the following:
trait Traversable1[B] {
def head: Option[B]
def next: Traversable1[B]
}
This works well, except I want next to return an object that has the same type as the original traversable, not the more generic traversable type. For example, if List has the trait Traversable1, it's next method should return something of type List, not Traversable1. Thus, my second attempt was the following:
trait Traversable2[A[B] <: Traversable2[A[B]] {
def head: Option[B]
def next: A[B]
}
Here, A could equal List, so next would return a List[B]. Unfortunately, this code does not compile: A[B] takes no type parameters, expected: one
How can I achieve what I'm trying to do?

Answer 1

You could create a trait like:
trait Traversable[T[_]] {
def head[A](t: T[A]): Option[A]
def next[A](t: T[A]): T[A]
}
then implement it for lists like:
implicit val listTraversable = new Traversable[List] {
def head[A](l: List[A]) = l match {
case Nil => None
case x::_ => Some(x)
}
def next[A](l: List[A]) = l.tail
}
You can then make use of it using the 'type class' pattern by taking an implicit implementation of the trait to do the work e.g.
def foreach[T[_], A](t: T[A], f: A => Unit)(implicit trav: Traversable[T]): Unit = {
trav.head(t) match {
case Some(v) =>
f(v)
foreach(trav.next(t), f)
case None => ()
}
}
and call it with
foreach(List(1,2,3), println)