Ex. 3.5 in ACL

[Robert L.]

Jock Cooper wrote:

> > I am just trying to solve Ex 3.5 in Graham's ANSI Common Lisp book. I
> > am reading it on my own and not as part of a university course. The
> > task is:
> >
> > define a function pos+, that takes a list as param and returns a list
> > that adds the position of each element of the list to the element's
> > value.
> > Thus:
> > (pos+ '(7 5 1 4))
> > returns:
> > (7 6 3 7)
> >
> > This is easy to solve using recursion, but the function shall also be
> > defined using iteration and using mapcar.
> >
> > My solutions are as follows, but they horribly and senselessly use
> > side-effects and thus look very inelegant. Is there a better way to do
> > it? (I mean, using just very primitive means - this exercise is in the
> > very beginning of the book...)
> >
> > Thanks!
> >
> > Chris
> >
> > mapcar:
> >
> > (defun pos+1 (lst)
> > (setf i 0)
> > (mapcar #'(lambda (x) (let ((new (+ x i)))
> > (progn (setf i (+ i 1))
> > new)))
> > lst))
> >
>
> You're indeed right that some of your side effects are unneeded.
>
> (defun pos+1 (lst &aux (pos -1))
> (mapcar #'(lambda (val) (+ val (incf pos))) lst))
>
> Most people don't like using &aux; I think for a short function like
> this it's OK.
>
>
> > iteration:
> >
> > (defun pos+ (lst)
> > (setf acc NIL)
> > (setf i 0)
> > (dolist (obj lst)
> > ; i know, instead of append, i could do a cons and reverse afterwards...
> > (progn (setf acc (append acc (list (+ obj i))))
> > (setf i (+ i 1))))
> > acc)
>
> (defun pos+ (lst)
> (do* ((pos 0 (1+ pos))
> (lst-cdr lst (cdr lst-cdr))
> (lst-car #1=(car lst-cdr) #1#)
> result)
> ((null lst-cdr) (nreverse result))
> (push (+ pos lst-car) result)))
>
> my preferred way would use LOOP:
>
> (defun pos+ (lst)
> (loop for val in lst
> for pos upfrom 0
> collect (+ val pos)))

(require srfi/42) ; list-ec

(define (pos+ lst)
(list-ec (:list x (index i) lst) (+ x i)))

(pos+ '(200 300 400 500))
===>
'(200 301 402 503)

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