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

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.

Came across this old thread while attempting to do this same thing. I eventually settled on this solution:

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

val testFoo = Foo("a string", 1)

val (str, in) = testFoo match { case Foo(f, b) => (f, b) }