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Home/ Questions/Q 7685521
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T19:16:10+00:00 2026-05-31T19:16:10+00:00

Here is what I have so far : def mergesort[T <: Ordered[T]](elements : List[T])

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Here is what I have so far :

  def mergesort[T <: Ordered[T]](elements : List[T]) : List[T] = {
    def merge(first : List[T], second : List[T]) : List[T] = (first, second) match {
      case (Nil, _) => second
      case (_, Nil) => first
      case (x :: xs, y :: ys) => if (x < y) x :: merge(xs, second) else y :: merge(first, ys)
    }

    if (elements.isEmpty) Nil
    else {
      val length = elements.length
      val (firstHalf, secondHalf) = elements.splitAt(length/2)

      merge(mergesort(firstHalf), mergesort(secondHalf))
    }
  }

The problem I’m having is that this fails

mergesort(List(1, 3, 6, 3, 1, 0))

error: inferred type arguments [Int] do not conform to method mergesort's type parameter bounds [T <: Ordered[T]]
       mergesort(List(1, 3, 6, 3, 1, 0))
       ^

Is there any way to make this work for all types which are ordered? I though Scala would have some sort of implicit conversion to a ‘rich’ integer type, which I assumed would have the Ordered trait.

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  1. Editorial Team
    Editorial Team
    2026-05-31T19:16:11+00:00Added an answer on May 31, 2026 at 7:16 pm

    What you need is a view bound def mergesort[T <% Ordered[T]]. See answers to this question: Generic method to return first of two values.

    This will now complie but you have some bugs in your algorithm giving you a stackoverflow error in the splitAt line.

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