Instantiating a case class from a list of parameters

Question

Given:

case class Foo(a: Int, b: String, c: Double) 

you can say:

val params = Foo(1, "bar", 3.14).productIterator.toList 

and get:

params: List[Any] = List(1, bar, 3.14) 

Is there a way to "go backwards" and recreate a Foo object directly from this list, i.e.:

Foo.createFromList(params)   // hypothetical 

instead of writing:

Foo(params(0).asInstanceOf[Int], params(1).asInstanceOf[String], params(2).asInstanceOf[Double]) 

EDIT: it seems that it boils down to being able to send the elements of a list as parameters to a function without writing them out explicitly, e.g.:

def bar(a: Int, b: Int, c: Int) = //... val list = List(1, 2, 3, 4, 5) bar(list.take(3)) // hypothetical, instead of: bar(list(0), list(1), list(2)) 

I would sort of expect to be able to do:

bar(list.take(3): _*) 

but that doesn't seem to work.

EDIT: Solution based on extempore's answer, but invoking the constructor directly instead of using the apply method:

case class Foo(a: Int = 0, b: String = "bar", c: Double = 3.14) {     val cs = this.getClass.getConstructors     def createFromList(params: List[Any]) =     cs(0).newInstance(params map { _.asInstanceOf[AnyRef] } : _*).asInstanceOf[Foo] } 

Now you can do:

scala> Foo().createFromList(List(4, "foo", 9.81)) res13: Foo = Foo(4,foo,9.81) 

You can also refactor the creation method into a trait:

trait Creatable[T <: Creatable[T]] {     val cs = this.getClass.getConstructors     def createFromList(params: List[Any]) =         cs(0).newInstance(params map { _.asInstanceOf[AnyRef] } : _*).asInstanceOf[T]    }  case class Bar(a: Int = 0, b: String = "bar", c: Double = 3.14) extends Creatable[Bar] 

And do e.g.:

scala> val bar = Bar() bar: Bar = Bar(0,bar,3.14)  scala> bar == bar.createFromList(bar.productIterator.toList) res11: Boolean = true 

Answer 1

scala> case class Foo(a: Int, b: String, c: Double) defined class Foo  scala> val params = Foo(1, "bar", 3.14).productIterator.toList params: List[Any] = List(1, bar, 3.14)  scala> Foo.getClass.getMethods.find(x => x.getName == "apply" && x.isBridge).get.invoke(Foo, params map (_.asInstanceOf[AnyRef]): _*).asInstanceOf[Foo] res0: Foo = Foo(1,bar,3.14)  scala> Foo(1, "bar", 3.14) == res0 res1: Boolean = true 

Edit: by the way, the syntax so far only being danced around for supplying the tuple as an argument is:

scala> case class Foo(a: Int, b: String, c: Double) defined class Foo  scala> Foo.tupled((1, "bar", 3.14))                 res0: Foo = Foo(1,bar,3.14) 

Answer 2

You could use pattern matching like:

params match {                                     case List(x:Int, y:String, d:Double) => Foo(x,y,d) }