A generic quicksort in Scala

Question

Hi,

I've been playing around with Scala recently and was thinking about how to implement a generic version of quicksort in it (just to get a better feeling for the language)

I came up with something like this

object Main {
  def qs[T](a: List[T], f: (T, T) => Boolean): List[T] = { 
    if (a == Nil) return a
    val (l, g) = a drop 1 partition (f(a(0),(_:T)))
    qs(l, f) ::: List(a(0)) ::: qs(g, f)
  }

  def main(args: Array[String]): Unit = { 
    val a = List(5,3,2,1,7,8,9,4,6)
    val qsInt = qs(_: List[Int], (_: Int) > (_: Int))
    println(qsInt(a))
  }

}

This is not as generic as I wanted it to be, since I have to explicitly state how to order the elements rather then just doing something like

val (l, g) = a drop 1 partition (a(0) >)

How can I tell the compiler that T only needs to implement the greater-than operator to be sortable by this function?

Regards

Answer 1

def qsort[T <% Ordered[T]](list: List[T]): List[T] = {
  list match {
  case Nil => Nil     
  case x::xs =>        
    val (before, after) = xs partition (_ < x)
    qsort(before) ++ (x :: qsort(after))
  }
}

Answer 2

Since Roger covered the Ordered case, let me cover Ordering:

def qsort[T](list: List[T])(implicit ord: Ordering[T]): List[T] = list match {
  // import ord._ // enables "_ < x" syntax
  case Nil => Nil
  case x :: xs =>
    val (before, after) = xs partition (ord.lt(_, x))
    qsort(before) ::: x :: qsort(after)
}

Using Ordering has two main advantages:

1. The T type does not need to have been created as Ordered.
2. One can easily provide alternate orderings.

For instance, on Scala 2.8:

def sortIgnoreCase(strs: List[String]) = {
  val myOrdering = Ordering.fromLessThan { (x: String, y: String) => 
    x.toLowerCase < y.toLowerCase
  }
  qsort(strs)(myOrdering)
}