Question

I have the following items

(define itemslist
  (list 'a1 'b2 'c3 (list 'z1 'z2) 'd5 'e6))

My method to find items is below

(define find-item
  (lambda (item itemslist)
    (cond ((null? itemslist) #f)
          ((list? (car itemslist))
            (cond ((null? itemslist) #f)
                  (else (find-item item (car itemslist)))))
          ((equal? stn (car itemslist)) (display "found"))
          (else (find-stn stn (cdr itemslist)))
          )
    )
  )

With my method above I can find a1, b2, c3, z1, z2. But when I want to find d5 onwards, it returns nothing. It seems to have skip the stack. Btw, I am just starting to learn Scheme so simple explanation would be better.

One more qns, how about if I have this

(list 'a1 'b2 'c3 (list 'z1 (list 'y1 'y2) 'z2) 'd5 'e6)

does this works as well? Thanks!

Answer 1

Yes, you skip lists.
Example:

'(1 '(2 3) 4)

If (list? '(2 3)) is true, the else part (outer cond) wont be evaluated so 4 is skipped.
So, either you place the else part inside list? block or you redesign your code.

If the itemlist is a list/pair you can pass it right away in a recursive call so
you can avoid writing code like (car itemslist) inside predicates thus make the code simple.

Here is a redesigned (simplified) version:

(define (find-item item l)
  (cond ((equal? item l) #t)
        ((pair? l) (or (find-item item (car l)) (find-item item (cdr l))))
        (else #f)))

Tip: you can also can write lists with '(...) notation instead of (list ...), ie

(find-item 'z2 '('a1 'b2 'c3 '('z1 '('y1 'y2) 'z2) 'd5 'e6))  
#t  
(find-item 'e6 '('a1 'b2 'c3 '('z1 '('y1 'y2) 'z2) 'd5 'e6))  
#t