I am working in a Scala embedded DSL and macros are becoming a main tool for achieving my purposes. I am getting an error while trying to reuse a subtree from the incoming macro expression into the resulting one. The situation is quite complex, but (I hope) I have simplified it for its understanding.

Suppose we have this code:

val y = transform {
  val x = 3
  x
}
println(y) // prints 3

where 'transform' is the involved macro. Although it could seem it does absolutely nothing, it is really transforming the shown block into this expression:

3 match { case x => x }

It is done with this macro implementation:

def transform(c: Context)(block: c.Expr[Int]): c.Expr[Int] = {
  import c.universe._
  import definitions._

  block.tree match {
    /* {
     *   val xNam = xVal
     *   xExp
     * }
     */
    case Block(List(ValDef(_, xNam, _, xVal)), xExp) =>
      println("# " + showRaw(xExp)) // prints Ident(newTermName("x"))
      c.Expr(
        Match(
          xVal, 
          List(CaseDef(
            Bind(xNam, Ident(newTermName("_"))),
            EmptyTree,
            /* xExp */ Ident(newTermName("x")) ))))
    case _ => 
      c.error(c.enclosingPosition, "Can't transform block to function")
      block  // keep original expression
  }
}

Notice that xNam corresponds with the variable name, xVal corresponds with its associated value and finally xExp corresponds with the expression containing the variable. Well, if I print the xExp raw tree I get Ident(newTermName("x")), and that is exactly what is set in the case RHS. Since the expression could be modified (for instance x+2 instead of x), this is not a valid solution for me. What I want to do is to reuse the xExp tree (see the xExp comment) while altering the 'x' meaning (it is a definition in the input expression but will be a case L

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