sicp-ex-2.17

  
 ;; ex 2.17 
 (define (last-pair items) 
   (let ((rest (cdr items))) 
     (if (null? rest) 
         items 
         (last-pair rest)))) 
  
 ;; Test: 
 (last-pair (list 1 2 3 4)) 
  

 solution which doesn't fail for '() 
  
 (define (last-pair items) 
   (define (iter items result) 
     (if (null? items) 
         result 
         (iter (cdr items) items))) 
   (iter items items)) 

 A simple but ingenious code 
  
 (define(lastPair L) 
  (if(=(length L)1) (car L) 
     (lastPair (cdr L)))) 

 same as EcsCodes, but passes test for empty list 
  
 (define (last-pair items) 
  (if (< (length items) 2) items 
   (last-pair (cdr items)))) 

 ;; ex 2.17 
 (define(lastPair L)  
   (define (ref lst x) 
   (if(= x 0) (car lst) 
      (ref (cdr lst) (- x 1)))) 
 (if(=(length L)1) (car L)  
      (cons (ref  L (- (length_ L) 2)) 
            (cons (ref L (- (length_ L) 1)) '()) 
            ) 
      ) 
   ) 
  

The answer was meant to return the last item of a list as a list. So you must modify the return value to return a list.

 (define (last-pair l) 
   (cond ((null? l) l) 
         ((null? (cdr l)) (list (car l))) 
         (else (last-pair (cdr l))))) 
 ;; test 
 (last-pair (list 23 72 149 34)) ;;= (34) 
 (last-pair '()) ;;= () 

 (define (last-pair s) 
   (if (null? (cdr (cdr s))) 
     (cdr s) 
     (last-pair (cdr s)))) 

Regarding the first solution. We need to return a list, not a particular element. 1 != '(1)

 ;; returns the list that contains the last element  
 ;; of a given non-empty list 
  
  
 (define (last-pair L) 
   (if (null? (cdr L)) 
       L 
       (last-pair (cdr L)))) 
  
  
 (last-pair (list 1))  ;; '(1) 
 (last-pair (list 1 2))  ;; '(2) 
 (last-pair (list 1 2 3))  ;; '(3) 
 (last-pair (list 1 2 3 4))  ;; '(4)