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