Improper vs. proper list in Scheme

Question

The original code I try to implement.. the output should be (1.4) (2.5) from my code.. I think you all know what I try to do....this is also tail recursion practice

my code
(define (myFunc lst1 lst2)
 (if (or (null? lst1) (null? lst2))       
    '()                                  
     (list (cons (car lst1) (car lst2))
        (myFunc (cdr lst1) (cdr lst2)))
  ))

after several of you gave me good advice about cons-pair.. so now it get's the dotted symbol in the middle.. then problem is that the improper list with empty list in the end.. when 2 input lists are like this ..... '(1 2 3 4) '(4 5 6)) my output is like this ; ((1 . 4) ((2 . 5) ((3 . 6) ()))) the empty list in the end of output shouldn't be there... so I couldn't understand about improper list , proper list....? is there are any document, I can look at?

Answer 1

Consider the difference between cons and list:

That is, (cons a b) creates a cell whose car is a and cdr is b.
(list a b) creates a cell whose car is a, but the cdr is a list, and the car of that list is b, while its cdr is nil.

If b is a list, the one on the left will be a list which has b as its tail, and with a added at the front of b.
The one on the right will also be a list, but one which has b as its second element, not as its tail like you want.

To fix your program, you only need to replace your list with a cons.

But your function is not tail-recursive, because it does things with the result of the recursive call.

To make it tail-recursive, a good way is usually to make a helper function which has an accumulator parameter.

I would probably write it something like this:

(define (zip-cars l1 l2)
    (cons (car l1) (car l2)))

(define (zip-help l1 l2 result)
    (if (or (null? l1) (null? l2))
        result
        (zip-help (cdr l1) (cdr l2) (cons (zip-cars l1 l2) result))))

(define (zip l1 l2)
    (zip-help l1 l2 '()))