In Scala, is there an easy way to convert a case c

问题:

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.

回答1:

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()

回答2:

You might try extending the ProductN trait, for N=1-22, which TupleN extends. It will give you a lot of Tuple semantics, like the _1, _2, etc. methods. Depending on you how you use your types, this might be sufficient without creating an actual Tuple.

回答3:

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.

回答4:

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) }