Scala factory for generic types using the apply method?

Question

Suppose that I have the following trait that defines an interface and takes a couple of type parameters...

trait Foo[A, B] {

    // implementation details not important

}

I want to use the companion object as a factory for concrete implementations of the trait. I also want to force users to use the Foo interface instead of sub-classing So I hide the concrete implementations in the companion object like so:

object Foo {

  def apply[A, B](thing: Thing): Foo[A, B] = {
    ???
  }

  private case class FooImpl[A1, B1](thing: Thing) extends Foo[A1, B1]

  private case class AnotherFooImpl[A2, B1](thing: Thing) extends Foo[A2, B1]

}

I want to be able to use the factory as follows:

val foo = Foo[A1, B1](thing)  // should be an instance of FooImpl

val anotherFoo = Foo[A2, B1](thing)  // should be an instance of AnotherFooImpl

How do I implement the apply method to make this happen? This SO post seems close to the mark.

Answer 1

How about:

trait Foo[A, B]
trait Factory[A, B] {
  def make(thing: Thing): Foo[A, B]
}

class Thing

object Foo {
def apply[A, B](thing: Thing)(implicit ev: Factory[A, B]) = ev.make(thing)

private case class FooImpl[A, B](thing: Thing) extends Foo[A, B]
private case class AnotherFooImpl[A, B](thing: Thing) extends Foo[A, B]

implicit val fooImplFactory: Factory[Int, String] = new Factory[Int, String] {
  override def make(thing: Thing): Foo[Int, String] = new FooImpl[Int, String](thing)
}

implicit val anotherFooImplFactory: Factory[String, String] = new Factory[String, String] {
  override def make(thing: Thing): Foo[String, String] = new AnotherFooImpl[String, String](thing)
}

And now:

def main(args: Array[String]): Unit = {
  import Foo._

  val fooImpl = Foo[Int, String](new Thing)
  val anotherFooImpl = Foo[String, String](new Thing)

  println(fooImpl)
  println(anotherFooImpl)
}

Yields:

FooImpl(testing.X$Thing@4678c730)
AnotherFooImpl(testing.X$Thing@c038203)