I am trying to solve a (possibly) trivial problem. I would like a nice, concise way to instantiate Array of bytes based on range. So far this works
Array(1 : Byte, 2 : Byte)
but I would like to use sth like
((1: Byte) to (10: Byte)).toArray
this is however Array[Int].
Range is not generic; it inherits from IndexedSeq[Int], so there's no way to make a "Range of Byte". (Edit: See Daniel C. Sobral's answer for a generic range type!)
When you try ((1: Byte) to (10: Byte)), the Bytes are implicitly converted back to Int again.
How about:
(1 to 10).map(_.toByte).toArray
That will result in two passes over the collection; if that's an issue, a non-strict view will rectify that:
(1 to 10).view.map(_.toByte).toArray
While Ben James answer is true enough, there is a more generic range for any type T for which there is an Intergral[T]: NumericRange.
import scala.collection.immutable.NumericRange
NumericRange(1: Byte, 10:Byte, 1: Byte).toArray
Another alternative would be mapping the resulting array to byte instead of mapping the range. For example, and using an Array method:
Array.range(1, 10).map(_.toByte)
NumericRange as I was writing that answer, but didn't investigate enough!