CODEWITHSUNDEEP
listschemelispconditional-statements
user3241846
user3241846

Reputation: 677

cond with list in Scheme

problem takes a list e.g L = (4 11 16 22 75 34) and gets tested for a condition (modulus 2) then returns a list with all the items in the list that pass the test e.g newL = (4 16 34)

Here is the code:

(define clean-list
  (lambda (x)
    (list x (test x))))

(define test
  (lambda (x)
    (cond (= 0 (modulo (car x) 2)) 
          (cons (car x) (test(cdr x)))   
          (else (test(cdr x))))
    ))

output:

   ((4 11 16 22 75 34) 0)

I debugged the code, it foes into (modulo (car x) 2) then returns to clean-list and exits, all after first run, please explain that and why it returns a 0 at the end of list. Also any feedback or improvement in code would be appreciated.

Upvotes: 0

Views: 701

Answers (2)

R Sahu
R Sahu

Reputation: 206567

Here's a tail-recursive version of the function.

(define test
  (lambda (x)

    (define (helper in out)
      (if (null? in)
        out
        (if (= 0 (modulo (car in) 2))
          (helper (cdr in) (append out (list (car in))))
          (helper (cdr in) out))))

    (helper x '())
    ))

(define clean-list
  (lambda (x)
    (list x (test x))))

(write (clean-list '(4 11 16 22 75 34)))

The output:

((4 11 16 22 75 34) (4 16 22 34))

PS. When I tested the code at repl.it, I had to change modulo to mod.

Upvotes: 0

Barmar
Barmar

Reputation: 780798

You're missing a set of parentheses. You're also missing a test for the recursion bottoming out.

(define test
  (lambda (x)
    (cond ((eq? x '()) '())
          ((= 0 (modulo (car x) 2)) 
           (cons (car x) (test(cdr x))))
          (else (test(cdr x))))
    ))

DEMO

The general syntax of cond is:

(cond (<test1> <result1>)
      (<test2> <result2>)
      ...)

In your code <test1> was simply =, not (= 0 (modulo (car x) 2)).

Upvotes: 3

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