Scheme: lambda return #<void> instead nothing

Question

i want to implement a lazy list that get 2 numbers as argument: low & int. The lambda will returns the lazy-list of all integers that are larger than low and divisible by int.

For example:

> (define lz1 (div-from 5 12))
> (take lz1 3)
'(12 24 36)

My try of implementation:

>(define gen_item
       (lambda (n int)
           (cons (cond ((= 0 (modulo n int)) n)) 
               (lambda () (gen_item (+  n 1) int)))))

When using this take implemenatation:

>(define take
 (lambda (lz-lst n)
 (if (= n 0)
 (list)
 (cons (car lz-lst)
 (take (tail lz-lst) (sub1 n))))))

When i run the lambda:

(take (gen_item 5 12) 20)

The return value:

'(#<void>      #<void>      #<void>      #<void>      #<void>
  #<void>      #<void>      12      #<void>      #<void>
  #<void>      #<void>      #<void>      #<void>      #<void>
  #<void>      #<void>      #<void>      #<void>      24)

How can i prevent the lambda return #<void> and return nothing instead?

Thank you.

Answer 1

your cond does not have a default case. So when (modulo n int) is not zero your get a undefined value. Add one like this:

(cond (predicat-expression consequent-expression)
      (predicat-expression2 consequent-expression2)
      (else alternative-expression))

Since you only have one consequent expression you could do this with if

(if predicate-expression
    consequent-expression
    alternative-expression)

Now if you don't any elements when predicate is not true then yuu should not cons anything when it's false but continue to the next true value:

(define gen_item 
  (lambda (n int)
    (if (= 0 (modulo n int))
        (cons n (lambda () (gen_item (+  n 1) int)))
        (gen_item (+  n 1) int))))