scala macro generic field of generic class not apply class generic type parameter

Question

Generic case class

case class GroupResult[T](
  group: String,
  reduction: Seq[T]
)

Macro method

 def foo[T] = macro fooImpl[T]

 def fooImpl[T: c.WeakTypeTag](c: Context) = {
    import c.universe._
    val tpe = weakTypeOf[T]
     tpe.declarations.collect {
      case m: MethodSymbol if m.isCaseAccessor => println(m.returnType)
    }
    c.literalUnit
  }

When I invoke foo[GroupResult[Int]]

The output is

String
Seq[T]

T is not applied ? How can I get the applied Seq[Int] ?

Answer 1

You can use typeSignatureIn to get the type signature of a method given GroupResult[Int]:

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

case class GroupResult[T](group: String, reduction: Seq[T])

def foo[T] = macro fooImpl[T]

def fooImpl[T: c.WeakTypeTag](c: Context) = {
  import c.universe._

  val tpe = weakTypeOf[T]

  tpe.declarations.collect {
    case m: MethodSymbol if m.isCaseAccessor => println(m.typeSignatureIn(tpe))
  }

  c.literalUnit
}

And then:

scala> foo[GroupResult[Int]]
=> String
=> Seq[Int]

So we're closer, but now we're getting the "types" of the accessors, not their return types. If we want the return types we can use the NullaryMethodType extractor:

def foo[T] = macro fooImpl[T]

def fooImpl[T: c.WeakTypeTag](c: Context) = {
  import c.universe._

  val tpe = weakTypeOf[T]

  tpe.declarations.collect {
    case m: MethodSymbol if m.isCaseAccessor => m.typeSignatureIn(tpe) match {
      case NullaryMethodType(returnType) => println(returnType)
    }
  }

  c.literalUnit
}

And then:

scala> foo[GroupResult[Int]]
String
Seq[Int]

And we're done.