CODEWITHSUNDEEP
scalaimplicit
Dranna
Dranna

Reputation: 505

Scala implicit constructor

Is it possible, in Scala, to instanciate an object without actually calling its name ?

In particular, I have :

val foo = Thing(
  Thingy(1,2,3),
  Thingy(4,5,6)
)

And I wonder if it would be possible to call it like that

val foo = Thing(
  (1,2,3),
  (4,5,6)
)

Upvotes: 3

Views: 373

Answers (2)

Krzysztof Atłasik
Krzysztof Atłasik

Reputation: 22595

You can use implicit conversion from Tuple3 to Thingy:

package example

case class Thingy(v1:Int, v2:Int, v3:Int)

object Thingy {
   implicit def tuple2Thingy(t: Tuple3[Int, Int, Int]) = Thingy(t._1, t._2, t._3)
   //add conversion in companion object
}

then you can use it like this:

import example.Thingy._

val foo = Thing(
    (1,2,3),
    (4,5,6)
)

If Thingy is vararg:

case class Thingy(v1:Int*)

object Thingy {
   implicit def tuple2ToThingy(t: Tuple2[Int, Int]) = Thingy(t._1, t._2)
   implicit def tuple3ToThingy(t: Tuple3[Int, Int, Int]) = Thingy(t._1, t._2, t._3)
   //etc until Tuple22
}

Upvotes: 2

Leo C
Leo C

Reputation: 22439

Here's another approach:

case class Thingy(a: Int, b: Int, c: Int)

case class Thing(x: Thingy, y: Thingy)

object Thing {
   def apply(t1: Tuple3[Int, Int, Int], t2: Tuple3[Int, Int, Int]): Thing =
     Thing(Thingy(t1._1, t1._2, t1._3), Thingy(t2._1, t2._2, t2._3))
}

Upvotes: 4

Related Questions