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
         


        

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)
}

Answer 3

Well, you can certainly do this with a tuple:

(Foo _).tupled apply (1, bar, 3.14)

But there is no real way to get from a List[S] to (A, B, C) for A, B, C <: S. There may be a way of doing this with HLists of course

Answer 4

Another one liner using case class companion object curried method and completely ignoring type safety :)

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

scala> val lst = List(1, "bar", 3.14)
lst: List[Any] = List(1, bar, 3.14)

scala> val foo = lst.foldLeft(Foo.curried: Any){case (r, v) => r.asInstanceOf[Function[Any, _]](v) }.asInstanceOf[Foo]
foo: Foo = Foo(1,bar,3.14)