Convert expression to polish notation in Scala - scala

Question

I would like to convert an expression such as: a.meth(b) to a function of type (A, B) => C that performs that exact computation.
My best attempt so far was along these lines:
def polish[A, B, C](symb: String): (A, B) => C = { (a, b) =>
// reflectively check if "symb" is a method defined on a
// if so, reflectively call symb, passing b
}
And then use it like this:
def flip[A, B, C](f : (A, B) => C): (B, A) => C = {(b, a) => f(a,b)}
val op = flip(polish("::"))
def reverse[A](l: List[A]): List[A] = l reduceLeft op
As you can pretty much see, it is quite ugly and you have to do a lot of type checking "manually".
Is there an alternative ?

Answer 1

You can achieve it easily with plain old subtype polymorphism. Just declare interface
trait Iface[B, C] {
def meth(b: B): C
}
Then you could implement polish easily
def polish[B, C](f: (Iface[B, C], B) => C): (Iface[B, C], B) => C = { (a, b) =>
f(a, b)
}
Using it is completely typesafe
object IfaceImpl extends Iface[String, String] {
override def meth(b: String): String = b.reverse
}
polish((a: Iface[String, String], b: String) => a meth b)(IfaceImpl, "hello")
Update:
Actually, you could achieve it using closures only
def polish[A, B, C](f: (A, B) => C): (A, B) => C = f
class Foo {
def meth(b: String): String = b.reverse
}
polish((_: Foo) meth (_: String))(new Foo, "hello")
Or without helper function at all :)
val polish = identity _ // Magic at work
((_: Foo) meth (_: String))(new Foo, "hello")