In Scala, is there an easy way to convert a case class into a tuple?

Is there an easy way to convert a case class into a tuple?

I can, of course, easily write boilerplate code to do this, but I mean without the boilerplate.

What I'm really after is a way to easily make a case class lexicographically Ordered. I can achieve the goal for tuples by importing scala.math.Ordering.Implicits._, and voila, my tuples have an Ordering defined for them. But the implicits in scala.math.Ordering don't work for case classes in general.

How about calling unapply().get in the companion object?

case class Foo(foo: String, bar: Int)

val (str, in) = Foo.unapply(Foo("test", 123)).get
// str: String = test
// in: Int = 123

Shapeless will do this for you.

  import shapeless._
  import shapeless.syntax.std.product._

  case class Fnord(a: Int, b: String)

  List(Fnord(1, "z - last"), Fnord(1, "a - first")).sortBy(_.productElements.tupled)

Gets

res0: List[Fnord] = List(Fnord(1,a - first), Fnord(1,z - last))

productElements turns a case class into a Shapeless HList:

scala> Fnord(1, "z - last").productElements
res1: Int :: String :: shapeless.HNil = 1 :: z - last :: HNil

And HLists are converted to tuples with #tupled:

scala> Fnord(1, "z - last").productElements.tupled
res2: (Int, String) = (1,z - last)

Performance is likely to be horrible, since you're constantly converting. You'd probably convert everything to the tupled form, sort that, then convert it back using something like (Fnord.apply _).tupled.