variable length arguments with generic types scala

Question

I have a simple abstract data type, a tree.

sealed trait Tree[A]
case object EmptyTree extends Tree[Nothing]
case class Leaf[A](value: A) extends Tree[A]
case class Node[A](op: A => A, branches: List[Tree[A]]) extends Tree[A]

How can i make the operation that a Node holds to accept a variable length number of arguments of type A?

An exemple:

def printAll(strings: String*) {
  strings.foreach(println)
}

printAll receives a variable number of strings.

In the same way i wanted to make my Tree to hold an operation op of the type A* => A.

A possible way would be to make it receive a List[A], but i wonder if there isn't a more direct way of doing this.

by the way can i also turn branches: List[Tree[A]] into something like branches: Tree[A]*. Does this make sense from a design perspective?

Mikel San Vicente · Accepted Answer · 2016-12-14 00:13:09Z

3

You can use Seq in the signature, then you will be able to pass varargs functions

sealed trait Tree[A]
case object EmptyTree extends Tree[Nothing]
case class Leaf[A](value: A) extends Tree[A]
case class Node[A](op: Seq[A] => A, branches: List[Tree[A]]) extends Tree[A]

object Test {
  Node[String](test[String], List())

  def test[A](elem: A*): A = ???
}

edited Dec 14, 2016 at 0:13

answered Dec 13, 2016 at 23:49

Mikel San Vicente

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3 Comments

raul ferreira Over a year ago

going to test it, it makes sense, than you for your answer

raul ferreira Over a year ago

by the way, can i also make branches: List[Tree[A]]) like branches: Tree[A]*. Does this make sense from a design perspective?

Mikel San Vicente Over a year ago

The varargs makes sense when you want to have a function that accepts n elements as parameter, but it makes less sense when you are declaring the fields of a class. But it's up to you to use it.

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