How to create an instance of a generic type in a Scala 3 macro?

Question

I'm porting a macro from Scala 2 to Scala 3. As part of it's work, the Scala 2 macro creates an instance of a generic type using the default constructor. This is simple to do with a quasiquote in Scala 2, but I'm struggling with Scala 3 macros. This is my best approach so far:

import scala.quoted.*

inline def make[A <: AnyRef]: A = ${ makeThat[A] }

private def makeThat[A <: AnyRef : Type](using Quotes): Expr[A] =
  import quotes.reflect.*

  '{ new A().asInstanceOf[A] }

Without the .asInstanceOf[A], the compiler emits an error message:

[error] -- [E007] Type Mismatch Error: ...
[error] 17 |  '{ new A() }
[error]    |     ^^^^^^^
[error]    |Found:    Object
[error]    |Required: A
[error]    |
[error]    |where:    A is a type in method makeThat with bounds <: AnyRef
[error] one error found

Is there a better solution without a downcast at runtime?

EDIT: As of Scala 3.0.1, this doesn't even compile anymore.

Answer 1

You can create the right tree with the lower level reflect API.

import scala.quoted.*

inline def make[A <: AnyRef]: A = ${ makeThat[A] }

def makeThat[A <: AnyRef : Type](using Quotes): Expr[A] =
  import quotes.reflect.*

  TypeRepr.of[A].classSymbol.map( sym =>
    Apply(Select(New(TypeTree.of[A]), sym.primaryConstructor), Nil).asExprOf[A]
  )
  .getOrElse(???) // not a class, so can't instantiate

Though you should include a check to see if your constructor doesn't have parameters.