Call new on class with type parameters

[gregghz]

Given this class:

class Thing[T](x: String)

I want to write a macro such that when it's called like this:

val x = foo

It produces the equivalent of this:

val x = new Thing[Int]("hello")

This is my attempt:

@experimental
inline def foo: Any = ${ impl }
@experimental
def impl(using Quotes): Expr[Any] = {
  import quotes.reflect.*

  val sym = Symbol.classSymbol("Thing")

  val t = Applied(TypeTree.ref(sym), List(TypeTree.of[Int]))
  val res = Apply(
    Select(
      New(t),
      sym.primaryConstructor
    ),
    List('{ "hello" }.asTerm)
  ).asExpr

  println(res.show)

  res
}

The println prints out:

new Thing[scala.Int]("hello")

but the compiler throws the following error:

[error] -- Error: 
[error]    |    val x = foo
[error]    |            ^^^
[error]    |            constructor Thing in class Thing does not take parameters

I did double check and without type params it does work fine (assuming the class is modified to have no type params):

val res = Apply(
      Select(
        New(TypeIdent(sym)),
        sym.primaryConstructor
      ),
      List('{ "hello" }.asTerm)
    ).asExpr

How can I successfully call the constructor while passing in type parameters?

[gregghz]

    @experimental
    private def newCall(
        cls: Symbol,
        typeArgs: List[TypeRepr],
        valArgs: List[Term]
    ): Apply = {
      New(Inferred(cls.typeRef))
        .select(cls.primaryConstructor)
        .appliedToTypes(typeArgs)
        .appliedToArgs(valArgs)
    }

This produces a valid new call to whatever class cls references with the given type and value parameters.