infix operator with automatic number conversion

Question

This simplified (and somewhat contrived) example is pretty close to what I'm looking for.

implicit class PlusMinus(a: Double) {
  def +-(b: Double) = if (a > b) a-b else a+b
}

With this I can:

scala> 3 +- 4L
res0: Double = 7.0

scala> 5f +- 1
res1: Double = 4.0

scala> 7L +- 6f
res3: Double = 1.0

But I have to wonder:

1. Every result is a Double. Can I imitate the automatic number conversions in the standard library?

   Int +- Int = Int
   Long +- Long = Long
   Long +- Float = Float  // etc.
   

2. Is there a better way? (There's always a better way.)

Answer 1

You could implement your operator like this

 implicit class PlusMinus[T](a: T)(implicit ev : Numeric[T])  {
    def +-(b: T) = if (ev.gt(a, b)) ev.minus(a, b)  else ev.plus(a  , b)
 }

This quick solution has a problem, it'ok only with the same type in the two operands.

This one seems to respond to your problem :

        implicit class PlusMinus[T](a: T)(implicit ev : Numeric[T])  {

        def +-(b: Double) = {
            val ad = ev.toDouble(a)
            if (ad > b) ad - b  else ad + b
        }

        def +-(b: Long) = {
            val ad = ev.toLong(a)
            if (ad > b) ad - b  else ad + b
        }

        def +-(b: Int) = {
            val ad = ev.toInt(a)
            if (ad > b) ad - b  else ad + b
        }

        def +-(b: Float) = {
            val ad = ev.toFloat(a)
            if (ad > b) ad - b  else ad + b
        }

        }

We have that result

3 +- 4                                    //> res0: Int = 7
3 +- 4L                                   //> res1: Long = 7
3L +- 4                                   //> res2: Int = 7

You can see in the last one, that the result type is the type of the second operand.