Can i fill function named parameters from seq in scala?

Question

If I have Seq('a -> 1, 'b -> "some text"), how can I call a function f(a: Int, b: String), filled with parameters from Seq, bound by name (a, b)?

Possible duplicate of http://stackoverflow.com/questions/20731069/unwrapping-a-list-or-map-as-function-arguments-in-scala — mattsilver, Apr 28 '15 at 06:44
Seq may be Seq('a -> 1, 'b -> "some text"), or Seq('b -> "some text", 'a -> 1), so varargs and tupled are not approach. From post (how i understand): "However, there is no analogous Scala syntax." — andrey.ladniy, Apr 28 '15 at 06:53

score 0 · Answer 1 · answered Apr 28 '15 at 07:28

You should be able to do it using scala-reflect, I believe. Something like this (untested):

def applyWithNamedParameters[T: ClassTag](obj: T, m: MethodSymbol, params: Map[String, Any]) {
  val methodMirror = scala.universe.runtime.currentMirror.reflect(obj).reflectMethod(m)
  val paramNames = m.paramss.flatten.map(_.name.toString)
  val paramValues = paramNames.map(params)

  methodMirror.apply(paramValues)
}

params are given as a map instead of Seq to simplify code a bit.

Can i fill function named parameters from seq in scala?

1 Answers1