use new wildcard type syntax

Author: Jasper-M

library/src/scala/quoted/ToExpr.scala

@@ -81,7 +81,7 @@ object ToExpr {
   }
 
   /** Default implemetation of `ToExpr[Class[T]]` */
-  given ClassToExpr[T <: Class[_]]: ToExpr[T] with {
+  given ClassToExpr[T <: Class[?]]: ToExpr[T] with {
     def apply(x: T)(using Quotes) = {
       import quotes.reflect._
       Ref(defn.Predef_classOf).appliedToType(TypeRepr.typeConstructorOf(x)).asExpr.asInstanceOf[Expr[T]]

tests/run-macros/quoted-toExprOfClass/Macros_1.scala

@@ -1,10 +1,10 @@
 import scala.quoted._
 
-inline def wildcard: Map[String, Class[_]] = ${ wildcardMacro }
+inline def wildcard: Map[String, Class[?]] = ${ wildcardMacro }
 inline def noWildcard: Map[String, Class[Int]] = ${ noWildcardMacro }
 
-def wildcardMacro(using Quotes): Expr[Map[String, Class[_]]] = {
-  val result: Map[String, Class[_]] = Map(
+def wildcardMacro(using Quotes): Expr[Map[String, Class[?]]] = {
+  val result: Map[String, Class[?]] = Map(
     "foo" -> classOf[Long],
     "bar" -> classOf[Int]
   )

tests/run-macros/quoted-toExprOfClass/Test_2.scala

@@ -1,6 +1,6 @@
 object Test {
   def main(args: Array[String]): Unit = {
-    val a: Map[String, Class[_]] = wildcard
+    val a: Map[String, Class[?]] = wildcard
     val b: Map[String, Class[Int]] = noWildcard
     println(a("foo"))
     println(a("bar"))