Question

I need a method to return the first of two ordered values. I’ve tried:

def first[T <: Ordered[T]](a: T, b: T) = {
  a compare b match {
    case -1 | 0 => a
    case 1      => b
  }
}

but get

scala> first(3,4)
<console>:9: error: inferred type arguments [Int] do not conform to method first's 
type parameter bounds [T <: Ordered[T]]
       first(3,4)
       ^

I guess this is because Int needs to be converted to a RichInt, which is an Ordered[Int] rather than an Ordered[RichInt]. What next?

Answer 1

You can use type class Ordering and context bound:

def first[T : Ordering](a: T, b: T) = {
  implicitly[Ordering[T]].compare(a, b) match {
    case -1 | 0 => a
    case 1      => b
  }
}

Update

This code can be simplified further if you import scala.math.Ordered._. Companion object of Ordered has orderingToOrdered implicit conversion, so everything that has Ordering would also be treated as Ordered:

import scala.math.Ordered._

def first[T : Ordering](a: T, b: T) = if (a <= b) a else b