[ Return the first and last element in the list Scheme ]

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.