Regression in tinkoff/oolong, missing type parameter in quotes

[WojciechMazur]

Compiler version

Works in 3.3.1-RC4
Fails in 3.3.2-RC nightly

Bisect points to 3c07379

Minimized code

import scala.quoted.*

case class QueryMeta[T](map: Map[String, String])

object QueryMeta:
  given [T]: FromExpr[QueryMeta[T]] = new FromExpr[QueryMeta[T]]:
    def unapply(expr: Expr[QueryMeta[T]])(using q: Quotes): Option[QueryMeta[T]] =
      import q.reflect.*
      expr match
        case '{ QueryMeta(${ map }: Map[String, String]) } =>
          map.value.map(QueryMeta[T].apply)
        case _ =>
          None

Output

-- Error: /Users/wmazur/projects/dotty/bisect/main.scala:10:16 -----------------
10 |        case '{ QueryMeta(${ map }: Map[String, String]) } =>
   |                ^^^^^^^^^
   |Reference to T within quotes requires a given scala.quoted.Type[T] in scope.
   |
1 error found

Workaround

        case '{ QueryMeta[t](${ map }: Map[String, String]) } =>

Expectation

It makes sense that quotes it fails when type parameter is failing, but the error message seems to not be very pricise here. It would make sense to make this code compile in 3.3.x for LTS source compatibility, and change semantics in 3.4.x

[nicolasstucki]

In 3c07379 we unintentionally fixed type inference for this pattern. Previously we wrongly inferred '{ QueryMeta[Any](...) } and now we infer '{ QueryMeta[T](...) }. Because we use this T in the quote pattern, we do need a given Type[T].

The correct fix for this code is

-  given [T]: FromExpr[QueryMeta[T]] = new FromExpr[QueryMeta[T]]:
+  given [T: Type]: FromExpr[QueryMeta[T]] = new FromExpr[QueryMeta[T]]:

Though the workaround is also fine

case '{ QueryMeta[t](${ map }: Map[String, String]) } =>