在Scala宏的方法调用中如何模型命名参数？

例如，让我们考虑以下案例类：

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

然后可以在每个这些case类实例上调用copy：

val newId = Some(1)

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

如here和here所述，没有简单的方法来抽象这样：

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)

所以我创建了一个scala宏，它做这个工作(几乎)：

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(...,...): " + currentType + "'")
    }

  }

}

Apply的最后一个参数(见上面代码块的底部)是一个参数列表(这里是“copy”方法的参数)。在新的宏API的帮助下，如何将类型为c.Expr [Option [Int])的给定id作为命名参数传递给复制方法？

特别是下面的宏表达式

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

应该导致

entity.copy(id = id)

以致如下

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)))

}

缺少的部分由占位符/ *？id？* /表示。