Converting BNF grammar to s-expression

Question

I defined a grammar called msl

#lang plai-typed
(define-type msl
  [msl-num (n : number)]
  [msl-add (l : msl) (r : msl)]
  [msl-mul (l : msl) (r : msl)]
  [msl-sub (l : msl) (r : msl)]
  [msl-pow (l : msl) (r : msl)]
  [msl-error (s : string)]
  )

(define(** u t)
       (cond
         ((= u 1) t)
         (else
          (* t(**(sub1 u) t)))))

I have a parser function which converts s-expression to msl

(define (parse [s : s-expression]) : msl
  (cond
    [(s-exp-number? s) (msl-num (s-exp->number s))]
    [(s-exp-list? s)
     (let ([sl (s-exp->list s)])
       (case (s-exp->symbol (first sl))
         [(+) (msl-add (parse (second sl)) (parse (third sl)))]
         [(*) (msl-mul (parse (second sl)) (parse (third sl)))]
         [ (-) (msl-sub (parse (second sl)) (parse (third sl)))]
         [ (**) (msl-pow (parse (second sl)) (parse (third sl))) ]
         [else (error 'parse "invalid list input")]))]
    [else (error 'parse "invalid input")]))

My question is: How can I convert msl expression to s-expression like this? I'm beginner on this subject

Answer 1

You can definitely do this.

Before I answer the question, though... why do you want to do this? It's not generally necessary in order to implement an evaluator.

With that said: it's fairly straightforward. I would begin by writing some test cases:

(test (unparse (msl-add (msl-num 3) (msl-num 4))) '(+ 3 4))

... and the the code itself

;; convert an msl to an s-expression:
(define (unparse parsed)
   (type-case msl parsed
      [msl-num (n) ...]
      ...))

... and so forth. Let me know if this doesn't make sense.