CODEWITHSUNDEEP
scalaschemeinterpreter
user1868607
user1868607

Reputation: 2610

Handling recursion in scala's scheme interpreter

I'm giving the following code that implements an interpreter of scheme language from scala. The relevant cases are those that contain a call to updateEnvRec or updateEnv:

 def evalRec(x: Data, env: Env): Data = {
    x match {
      case i: Int => i
      case Symbol(s) => env(s) match {
        case Some(v) => v
        case None => sys.error("Unknown symbol " + s)
      }
      case List('lambda, params: List[Data], body) =>
        ((args: List[Data]) => {
          val paramBinding = params.map(_.asInstanceOf[Symbol].name).zip(args)
          evalRec(body, updateEnv(env, paramBinding))
        })
      case List('val, Symbol(s), expr, rest) =>
        evalRec(rest, updateEnv(env, List(s -> evalRec(expr, env))))
      case List('def, Symbol(s), expr, rest) => {
        evalRec(rest, updateEnvRec(env, s, expr))
      }
      case List('if, bE, trueCase, falseCase) =>
        if (evalRec(bE, env) != 0) evalRec(trueCase, env)
        else evalRec(falseCase, env)
      case opE :: argsE => {
        val op = evalRec(opE, env).asInstanceOf[List[Data] => Data]
        val args: List[Data] = argsE.map((arg: Data) => evalRec(arg, env))
        op(args)
      }
    }
  }

The aim of this code is handle expressions such as:

(val factorial (lambda (x)
(if (= x 0) 1 (* x (factorial (- x 1)))))
(factorial 6))

To do so I'm given the following updates and definition of environment:

 type Env = String => Option[Data]
  val recEnv : Env = ((id:String) => funEnv.get(id)) 
  def updateEnv(env : Env, bindings : List[(String,Data)]) : Env = bindings match {
    case Nil => env
    case (id,d)::rest => ((x:String) =>
        if (x==id) Some(d)
        else updateEnv(env,rest)(x))
  }
  def updateEnvRec(env: Env, s : String, expr : Data) : Env = {
    def newEnv : Env = ((id:String) =>
      if (id==s) Some(evalRec(expr, newEnv))
      else env(id)
    )
    newEnv
  }

My problem is that I don't understand why calling:

 evalRec(rest, updateEnv(env, List(s -> evalRec(expr, env))))

works. I'd rather write:

evalRec(rest, updateEnvRec(env,s,expr))

because with the first strategy I would have to evaluate first evalRec(expr, env) and this would result in an error because at that point env does not contain factorial. What am I missing?

Upvotes: 2

Views: 193

Answers (1)

soegaard
soegaard

Reputation: 31145

I believe your example

(val factorial 
  (lambda (x)
    (if (= x 0) 
        1 
        (* x (factorial (- x 1)))))
(factorial 6))

is wrong. If I understand the code correctly, val doesn't introduce an recursive binding, therefore the the inner factorial is unbound. And you interpreter gives you just that error.

Instead try this program:

(def factorial 
  (lambda (x)
    (if (= x 0) 
        1 
        (* x (factorial (- x 1)))))
(factorial 6))

Upvotes: 1

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