Problem

If you have an object A that you want to convert to B.

Solution

Solution 1 - Define implicits

//Given:
case class A(value:String)
case class B(value:String)
def toB1(a:A):B = "B1-"+a.value
def toB2(a:A):B = "B2-"+a.value

//Reusable class
class Enriched[A](op: => A) {
	/** An object to an input object */
	def asRich: A = op
}

//Enrich them
object RichMapper {
  implicit def toB1(all: A) = new Enriched(toB1)
  implicit def toB2(all: A) = new Enriched(toB2)
}

//Use them
val a = A("a-value")
{
	import RichMapper.toB1
	val b = a.asRich
	println(b) // B1-a-value
}
{
	import RichMapper.toB2
	val b = a.asRich
	println(b) // B2-a-value
}
{
	import RichMapper.toB1
	val b = a.asRich //compilation error.
	println(b) // B1-a-value
}

Existing solutions

Use scala.collection.JavaConverters to interoperate with Java collections. These are a set of implicits that add asJava and asScala conversion methods. The use of these ensures that such conversions are explicit, aiding the reader:

import scala.collection.JavaConverters._

val list: java.util.List[Int] = Seq(1,2,3,4).asJava
val buffer: scala.collection.mutable.Buffer[Int] = list.asScala

References

http://stackoverflow.com/questions/14802517/is-there-a-way-to-enrich-a-scala-class-without-wrapping-the-code-into-another

Bad Solutions

TBD