Stack calculator in Racket

Question

I have implemented a stack calculator based on this programming task. I wondered if a more experienced racketeer could give me feedback and tell me if I am missing anything in Racket that would enable the solution to be more elegant.

#lang racket

(define (inc n)
  (+ n 1))

(define (print-value s)
  (begin
    (display s)
    (display " ")))

(define (exec l (stack empty) (opcount 1))
  "executes a stack based program."
  (cond
    [(empty? l) (void)]

    ;; push number onto stack
    [(number? (string->number (first l)))
     (exec (rest l)
           (cons (string->number (first l)) stack)
                 (inc opcount))]

    ;; pop number from stack
    [(equal? (first l) ".")
     (begin
       (print-value (first stack))
       (exec (rest l) (rest stack)
       (inc opcount)))]

    ;; mathmatical operators
    [(equal? (first l) "+")
     (exec (rest l) (cons
                     (+ (first stack) (second stack))
                     (rest (rest stack))) (inc opcount))]
    [(equal? (first l) "-")
     (exec (rest l) (cons
                     (- (first stack) (second stack))
                     (rest (rest stack))) (inc opcount))]
    [(equal? (first l) "*")
     (exec (rest l) (cons
                     (* (first stack) (second stack))
                     (rest (rest stack))) (inc opcount))]
    [(equal? (first l) "/")
     (exec (rest l) (cons
                     (/ (first stack) (second stack))
                     (rest (rest stack))) (inc opcount))]

    ;; duplication operator
    [(equal? (string-downcase (first l)) "dup")
     (exec (rest l)
           (cons (first stack) stack) (inc opcount))]
    [else
     (~a "Error: operation " opcount " invalid")]))

(define program (string-split "64 DUP * ."))

(exec program)

Answer 1

Bug

The - and / operators interpret their operands backwards from the typical RPN order. That is, I expect "10 7 -" should produce 3, but it actually produces -3.

Recursion

Your code is awfully repetitive, because the handler for each operator has to fetch the operator ((first l)), perform the operation, then make the recursive call ((exec (rest l) … (inc opcount))). It would be worthwhile to define a helper function that applies the operator to the stack.

(define (apply-op op stack)
  (cond
    [(number? (string->number op))
     (cons (string->number op) stack)]
    [(equal? op ".")
     (begin
       (display (first stack))
       (display " ")
       (rest stack))]
    [(equal? op "+")
     (cons (+ (second stack) (first stack)) (rest (rest stack)))]
    [(equal? op "-")
     (cons (- (second stack) (first stack)) (rest (rest stack)))]
    [(equal? op "*")
     (cons (* (second stack) (first stack)) (rest (rest stack)))]
    [(equal? op "/")
     (cons (/ (second stack) (first stack)) (rest (rest stack)))]
    [(equal? (string-downcase op) "dup")
     (cons (first stack) stack)]
    [else
     (~a "operation '" op "' invalid")]))

Then, the exec function just has to call it and drive the recursion.

(define (exec ops (stack empty) (opcount 1))
  "executes a stack based program."
  (cond
    [(empty? ops) (void)]
    [else
      (let ([next-stack (apply-op (first ops) stack)])
           (if (string? next-stack)
               (~a "Error at operator " opcount ": " next-stack)
               (exec (rest ops) next-stack (+ opcount 1))))]))