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copied!<p>The problem is that <code>Value</code> implemenents <code>Ordered[Enumeration#Value]</code>, rather than <code>Ordered[Dogs.Value]</code>. I don't know the rationale for this, it not be possible to do it the other way.</p> <p>This is sufficient for the simple case of directly comparing two values -- its just a normal method call:</p> <pre><code>scala> (Rover: Ordered[Enumeration#Value]).<(Sam) res44: Boolean = false </code></pre> <p>However, the type parameter <code>A</code> in <code>Ordered</code> is invariant, so when you you ask for type param to viewable as <code>Ordered[A]</code>, it is not sufficient to use <code>Dogs.Value <: Ordered[Enumeration#Value]</code>. If <code>A</code> were contra-variant this would be allowed, but it also causes other problems with type inference.</p> <p>You could work around the problem by statically typing the list with <code>Enumeration#Value</code>:</p> <pre><code>scala> val xs = List[Enumeration#Value](Rover, Tom, Sam, Sam, Rover) xs: List[Enumeration#Value] = List(Rover, Tom, Sam, Sam, Rover) scala> doSomething(xs) List(Sam, Sam, Tom, Rover, Rover) </code></pre> <p>Or, by explicitly passing the type parameter to <code>doSomething</code>:</p> <pre><code>scala> doSomething[Enumeration#Value](List(Rover, Sam)) List(Sam, Rover) </code></pre> <p>Or, better yet, loosening the requirements on the type parameters, essentially treating <code>Ordered</code> as contravariant in this circumstance.</p> <pre><code>scala> def doSomething[A <% Ordered[_ >: A]](xs : List[A]) = xs sortWith (_ < _) doSomething: [A](xs: List[A])(implicit evidence$1: (A) => Ordered[_ >: A])List[A] scala> doSomething(List(Rover, Sam)) res47: List[Dogs.Value] = List(Sam, Rover) </code></pre> <p>Why does that work?</p> <pre><code>scala> Rover: Ordered[_ <: Enumeration#Value] res52: scala.math.Ordered[_ <: Enumeration#Value] = Rover </code></pre>

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