Re: Lisp Function Problem

Write a Function 'total' that takes an orderd list ie.
((itemA quantityA costA)(itemB quantityB costB)....)
but returns a list giving the total cost plus the overall cost.
>
Eg:
>
LISP>(total'((book 2 10)(pen 3 2)(notepad 1 12)))
((BOOK 20)(PEN 6)(NOTEPAD 12)(TOTAL 38))
>
Thank you for your time
>

 
Here are 3 quite different solutions... one recursive, one iterative,
and one obscure. I won't do ALL your homework, so you will have to
figure out yourself how they work, which shouldn't be very hard.
Disclaimer: I'm quite new to Lisp myself, so there may be better and
more elegant solutions.
 
;;; 1
(defun total (lst)
   (labels ((total-rec (sublst subtot)
                       (if sublst
                         (let* ((sub (car sublst))
                                (sum (* (cadr sub) (caddr sub))))
                           (cons `(,(car sub) ,sum)
                               (total-rec (cdr sublst) (+ subtot sum))))
                         `((total ,subtot)))))
     (total-rec lst 0)))
 
;;; 2
(defun total (lst)
   (let ((total 0)
         (ret))
     (dolist (elt lst (nreverse (push `(total ,total) ret)))
       (let ((subtotal (* (cadr elt) (caddr elt))))
         (push `(,(car elt) ,subtotal) ret)
         (incf total subtotal)))))
 
;;; 3
(defun total (lst)
   (append
     (mapcar #'(lambda (e) (list (car e) (* (cadr e) (caddr e)))) lst)
     `((total ,(apply #'+ (mapcar #'(lambda (elt) (* (cadr elt) (caddr
elt))) lst))))))

Gauche Scheme

(define (proc stuff)
  (let* ((items (map car stuff))
         (costs (map (~>> cdr (apply *)) stuff))
         (total (fold + 0 costs)))
    (append (map list items costs)
            `((total ,total)))))

Given:

(define-syntax ->>
  (syntax-rules ()
    [(_ x)  x]
    [(_ x (y ...) z ...)
     (->> (y ... x) z ...)]
    [(_ x y z ...)
     (->> (y x) z ...)]))

(define-syntax ~>>
  (syntax-rules ()
    [(_ (func0 a ...) func ...)
     (lambda xs (->> (apply func0 a ... xs) func ...))]
    [(_ func0 func ...)
     (lambda xs (->> (apply func0 xs) func ...))]))