Cond interpreter Scheme

0
votes

I would like to write an interpreter for if and cond statements in scheme, however I couldnt do cond part.My cond has else part always.

<cond -> ( cond <conditional_list> <else_condition> )

<conditional_list -> <conditional | <conditional <conditional_list

<conditional -> ( <expr <expr )

<else_condition -> ( else <expr )

Here is my if part :

((if-stmt? e) 
        (if (not (eq? (my_interpreter (cadr e) env) 0))
                    ( my_interpreter (caddr e) env)
                    ( my_interpreter (cadddr e) env)))

Any help is appreciated thank you!

1
conditional-statementsschemeinterpreter
Write a separate function that evaluates a list of clauses. - molbdnilo
On a side note, those are expressions, not statements. - molbdnilo
Your interpreter should only handle the lowest level primitives (like if). Cond is better implemented as a macro. Same goes for things like let, let*, define, etc. - SK-logic

1 Answers

1
votes

As a comment says: don't handle cond: turn it into if. for instance (this may not get all of the semantics of cond correct in Scheme and is casual about errors):

(define (rewrite-cond cond-expr)
  (if (and (list? cond-expr)
           (eqv? (first cond-expr) 'cond))
      (rewrite-cond-clauses (rest cond-expr))
      (error 'rewrite-cond "not a cond")))

(define (rewrite-cond-clauses clauses)
  (if (null? clauses)
      '#f
      (let ((this (first clauses))
            (more (rest clauses)))
        (let ((test (first this))
              (body (rest this)))
          `(if ,(if (eqv? test 'else) '#t test)
             (begin ,@body)
             ,(rewrite-cond-clauses more))))))

Now

> (rewrite-cond '(cond ((x) 1 3)
                       (y 2)
                       (else 3)))
(if (x) (begin 1 3) (if y (begin 2) (if #t (begin 3) #f)))