Macro: is there a way to reify a spliced function?

[Léonard Schneider]

Hi,

I need to reify an expression of a spliced function. Unfortunately, reify does not work as expected when you append a wildcard. Instead of selecting the function which is being spliced it acts as if the spliced function were a value.

An example to illustrate the issue.

Welcome to Scala version 2.10.0-RC2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_37).
Type in expressions to have them evaluated.
Type :help for more information.
scala> import scala.reflect.runtime.{universe=>ru}
import scala.reflect.runtime.{universe=>ru}
scala> def f(x: Int) = 0
f: (x: Int)Int
scala> val g = ru.reify(f _)
g: reflect.runtime.universe.Expr[Int => Int] = 
Expr[Int => Int]({
  ((x) => f(x))
})
scala> ru.reify(g.splice _)
res0: reflect.runtime.universe.Expr[() => Int => Int] = 
Expr[() => Int => Int]((() => {
  ((x) => f(x))
}))

I would have expected an Expr[Int => Int]. Is this a bug? Is there a workaround?

Best regards,

Leo

[Eugene Burmako]

reify typechecks its arguments before doing any processing, so it
behaves exactly like normal Scala does:

scala> val g1 = f _
g1: Int => Int = <function1>

scala> val g2 = g1 _
g2: () => Int => Int = <function0>

Why it behaves like that is another question.

On Nov 15, 1:36 am, Léonard Schneider <leonard.schnei...@gmail.com>
wrote:

[Dave]

Maybe with parentheses to force apply after lifting otherwise it means only lifting.
Don't know if this is intended.

scala> val g2 = g1 _
g2: () => Int => Int = <function0>

scala> val g2 = g1(_)
g2: Int => Int = <function1>


scala> ru.reify(g.splice _)
res1: reflect.runtime.universe.Expr[() => Int => Int] =

Expr[() => Int => Int]((() => {
  ((x) => f(x))
}))

scala> ru.reify(g.splice(_))
res0: reflect.runtime.universe.Expr[Int => Int] =
Expr[Int => Int](((x$1) => {
  ((x) => f(x))
}.apply(x$1)))



Op donderdag 15 november 2012 10:38:29 UTC+1 schreef Eugene Burmako het volgende:

[Dave]

Or

scala> ru.reify((g.splice _).apply)
res5: reflect.runtime.universe.Expr[Int => Int] =
Expr[Int => Int]((() => {
  ((x) => f(x))
}).apply())


Op donderdag 15 november 2012 12:08:08 UTC+1 schreef Dave het volgende:

[Léonard Schneider]

Hi Dave,

I've tried that but it failed for my use case. So let me describe the more general problem I'm trying to solve. Any idea or hint is welcome.

// Challenge: write macro getf such as "def g = getf[A.type]" == "def g = A.f _" for any A object

// which contains a f method which can have any type signature

// This is just an example of an A object containing an f method

// object A {

//  def f(x: Int, s: String)

// }

def getf[T] = macro getfImpl[T]

def getfImpl[T: c.WeakTypeTag](c: Context) = {

  import c.universe._

  ???

}

[Léonard Schneider]

So here is a failed attempt to fix ideas. It shows why I've been trying to use the splice and wildcard combination in a reify. I think my initial question was a bad example to that regard.

scala> :pas
// Entering paste mode (ctrl-D to finish)
object A {
 def f(x: Int, s: String) = 0 

} 

def getf[T] = macro getfImpl[T]
def getfImpl[T: c.WeakTypeTag](c: Context) = {
  import c.universe._

  c.Expr(treeBuild.mkAttributedIdent(weakTypeOf[T].member(newTermName("f"))))
}
scala> def g = getf[A.type]
<console>:12: error: missing arguments for method f in object A;
follow this method with `_' if you want to treat it as a partially applied function

       def g = getf[A.type]

[Léonard Schneider]

OK. Answering myself. Here is what I came up with. It would still need to be made more robust by dealing with nullary methods for instance. In a few words, you have to redefine a function which takes the method values, as a lambda expression would, to be able to use it later on. I should be able to use the trick for automatically transform case classes into HList in my implementation ;)

object A {

 def f(x: Int, s: String) = 0 

} 

def getf[T]: Any = macro getfImpl[T]

def getfImpl[T: c.WeakTypeTag](c: Context) = {

  import c.universe._

  val fSym = weakTypeOf[T].member(newTermName("f")).asMethod 

  val argSyms = fSym.paramss.flatten

  val funTree = Function(argSyms.map(ValDef(_)), treeBuild.mkMethodCall(fSym, argSyms.map(s =>

    treeBuild.mkAttributedIdent(s)

  )))

  c.Expr(funTree)

[Dave]

Amazing. Well done!

C:\Users\Dave>scala
Welcome to Scala version 2.10.0-RC2 (Oracle JRockit(R), Java 1.6.0_37).

Type in expressions to have them evaluated.
Type :help for more information.

scala> import language.experimental.macros
import language.experimental.macros

scala> import scala.reflect.macros.Context
import scala.reflect.macros.Context

scala> :paste

// Entering paste mode (ctrl-D to finish)

object A {
 def f(x: Int, s: String) = 0
}

def getf[T]: Any = macro getfImpl[T]
def getfImpl[T: c.WeakTypeTag](c: Context) = {
  import c.universe._
  val fSym = weakTypeOf[T].member(newTermName("f")).asMethod
  val argSyms = fSym.paramss.flatten
  val funTree = Function(argSyms.map(ValDef(_)), treeBuild.mkMethodCall(fSym, argSyms.map(s =>
    treeBuild.mkAttributedIdent(s)
  )))
  c.Expr(funTree)
}

// Exiting paste mode, now interpreting.

defined module A
getf: [T]=> Any
getfImpl: [T](c: scala.reflect.macros.Context)(implicit evidence$1: c.WeakTypeTa
g[T])c.Expr[T]

scala> def g = getf[A.type]

g: (Int, String) => Int

Op zaterdag 17 november 2012 00:51:31 UTC+1 schreef Léonard Schneider het volgende: