Howto model named parameters in method invocations with Scala macros?

Question

There are use cases where it is useful to create a copy of an object which is an instance of a case class of a set of case classes, which have a specific value in common.

For example let's consider the following case classes:

case class Foo(id: Option[Int])
case class Bar(arg0: String, id: Option[Int])
case class Baz(arg0: Int, id: Option[Int], arg2: String)

Then copy can be called on each of these case class instances:

val newId = Some(1)

Foo(None).copy(id = newId)
Bar("bar", None).copy(id = newId)
Baz(42, None, "baz").copy(id = newId)

As described here and here there is no simple way to abstract this like this:

type Copyable[T] = { def copy(id: Option[Int]): T }

// THIS DOES *NOT* WORK FOR CASE CLASSES
def withId[T <: Copyable[T]](obj: T, newId: Option[Int]): T =
  obj.copy(id = newId)

So I created a scala macro, which does this job (almost):

import scala.reflect.macros.Context

object Entity {

  import scala.language.experimental.macros
  import scala.reflect.macros.Context

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

  def withIdImpl[T: c.WeakTypeTag](c: Context)(entity: c.Expr[T], id: c.Expr[Option[Int]]): c.Expr[T] = {

    import c.universe._

    val currentType = entity.actualType

    // reflection helpers
    def equals(that: Name, name: String) = that.encoded == name || that.decoded == name
    def hasName(name: String)(implicit method: MethodSymbol) = equals(method.name, name)
    def hasReturnType(`type`: Type)(implicit method: MethodSymbol) = method.typeSignature match {
      case MethodType(_, returnType) => `type` == returnType
    }
    def hasParameter(name: String, `type`: Type)(implicit method: MethodSymbol) = method.typeSignature match {
      case MethodType(params, _) => params.exists { param =>
        equals(param.name, name) && param.typeSignature == `type`
      }
    }

    // finding method entity.copy(id: Option[Int])
    currentType.members.find { symbol =>
      symbol.isMethod && {
        implicit val method = symbol.asMethod
        hasName("copy") && hasReturnType(currentType) && hasParameter("id", typeOf[Option[Int]])
      }
    } match {
      case Some(symbol) => {
        val method = symbol.asMethod
        val param = reify((
          c.Expr[String](Literal(Constant("id"))).splice,
          id.splice)).tree
        c.Expr(
          Apply(
            Select(
              reify(entity.splice).tree,
              newTermName("copy")),
            List( /*id.tree*/ )))
      }
      case None => c.abort(c.enclosingPosition, currentType + " needs method 'copy(..., id: Option[Int], ...): " + currentType + "'")
    }

  }

}

The last argument of Apply (see bottom of above code block) is a List of parameters (here: parameters of method 'copy'). How can the given id of type c.Expr[Option[Int]] be passed as named parameter to the copy method with the help of the new macro API?

In particular the following macro expression

c.Expr(
  Apply(
    Select(
      reify(entity.splice).tree,
      newTermName("copy")),
    List(/*?id?*/)))

should result in

entity.copy(id = id)

so that the following holds

case class Test(s: String, id: Option[Int] = None)

// has to be compiled by its own
object Test extends App {

  assert( Entity.withId(Test("scala rulz"), Some(1)) == Test("scala rulz", Some(1)))

}

The missing part is denoted by the placeholder /*?id?*/.

Answer 1

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.

Answer 2

This is the solution of Travis where all parts are put together:

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"))

    copy match {
      case s: MethodSymbol if (s.paramss.flatten.map(_.name).contains(
        newTermName("id")
      )) => c.Expr[T](
        Apply(
          Select(tree, copy),
          AssignOrNamedArg(Ident("id"), reify(id.splice).tree) :: Nil))
      case _ => c.abort(c.enclosingPosition, "No eligible copy method!")
    }

  }

}