Scala Doubles, and Precision

Question

Is there a function that can truncate or round a Double? At one point in my code I would like a number like: 1.23456789 to be rounded to 1.23

Answer 1

Recently, I faced similar problem and I solved it using following approach

def round(value: Either[Double, Float], places: Int) = {
  if (places < 0) 0
  else {
    val factor = Math.pow(10, places)
    value match {
      case Left(d) => (Math.round(d * factor) / factor)
      case Right(f) => (Math.round(f * factor) / factor)
    }
  }
}

def round(value: Double): Double = round(Left(value), 0)
def round(value: Double, places: Int): Double = round(Left(value), places)
def round(value: Float): Double = round(Right(value), 0)
def round(value: Float, places: Int): Double = round(Right(value), places)

I used this SO issue. I have couple of overloaded functions for both Float\Double and implicit\explicit options. Note that, you need to explicitly mention the return type in case of overloaded functions.