Howto model named parameters in method invocations with Scala macros?

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Here’s an implementation that’s also a little more generic:

import scala.language.experimental.macros

object WithIdExample {
  import scala.reflect.macros.Context

  def withId[T, I](entity: T, id: I): T = macro withIdImpl[T, I]

  def withIdImpl[T: c.WeakTypeTag, I: c.WeakTypeTag](c: Context)(
    entity: c.Expr[T], id: c.Expr[I]
  ): c.Expr[T] = {
    import c.universe._

    val tree = reify(entity.splice).tree
    val copy = entity.actualType.member(newTermName("copy"))

    val params = copy match {
      case s: MethodSymbol if (s.paramss.nonEmpty) => s.paramss.head
      case _ => c.abort(c.enclosingPosition, "No eligible copy method!")
    }

    c.Expr[T](Apply(
      Select(tree, copy),
      params.map {
        case p if p.name.decoded == "id" => reify(id.splice).tree
        case p => Select(tree, p.name)
      }
    ))
  }
}

It’ll work on any case class with a member named id, no matter what its type is:

scala> case class Bar(arg0: String, id: Option[Int])
defined class Bar

scala> case class Foo(x: Double, y: String, id: Int)
defined class Foo

scala> WithIdExample.withId(Bar("bar", None), Some(2))
res0: Bar = Bar(bar,Some(2))

scala> WithIdExample.withId(Foo(0.0, "foo", 1), 2)
res1: Foo = Foo(0.0,foo,2)

If the case class doesn’t have an id member, withId will compile—it just won’t do anything. If you want a compile error in that case, you can add an extra condition to the match on copy.

Edit: As Eugene Burmako just pointed out on Twitter, you can write this a little more naturally using AssignOrNamedArg at the end:

c.Expr[T](Apply(
  Select(tree, copy),
  AssignOrNamedArg(Ident("id"), reify(id.splice).tree) :: Nil
))

This version won’t compile if the case class doesn’t have an id member, but that’s more likely to be the desired behavior anyway.

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