Return the first and last element in the list Scheme

Question

Pretty straightforward question. My initial approach was to define another procedure to find the last element of lst within first-last. After finding the last element I appended it with the first element of lst (car lst). This is how append works. (append list1 list2) e.g., (append '(1 2 3) '(2 1 5)) -> (1 2 3 2 1 5) I'm wondering if the problem is just with my syntax but I am not sure.

(define (first-last lst)
 (define (last lst)
  (cond ((null? (cdr lst))(car lst))
  (else (last (cdr lst)))))
(append(car lst)(last lst)))

The error occurs in the

(append(car lst)(last lst)))

"mcar: contract violation

expected: mpair?

given: 1"

This is my first question on stack, so I'm sorry if the question is not presented in the correct way.

Answer 1

append is only for joining two or more lists. Here, though, you're not joining existing lists, but building a list from two elements. For that, use list:

(list (car lst) (last lst))

Answer 2

If you can use match, a neat solution is possible:

(define first-last
  (lambda (x)
    (match x
           ((first rest ... last)
            (list first last))
           ((only) (list only only))
           (_ #f))))

Of course, you could return something other than #f in the catch-all clause.