Re: doubling list elements at every level using recursion

On Apr 14, 8:53 pm, Slobodan Blazeski <slobodan.blaze...@xxxxxxxxx>
wrote:
On Apr 14, 8:43 am, pereges <Brol...@xxxxxxxxx> wrote:





thanks for all suggestions!

can someone also help me with a similar problem:

4.  Define an iterative function idouble that takes a list.  The
function idouble doubles every number at the top level in its argument
and does not alter other members of the argument.
Tests of idouble:

(idouble nil)                                           ; NIL
(idouble '(1 2 3))                              ; (2 4 6)
(idouble '(1 (2) 3))                            ; (2 (2) 6)
(idouble '(1 4 a e b 6))                         ; (2 8 A E B 12)

In this problem I cannot figure out how to iterate through a list
while at the same time modify its contents. This is the code I tried:

(defun idouble (lst)
(if (null lst)
NIL
(dolist (x lst)
 (if  (number p x)
(cons (* 2 x) (cdr lst))
(cons x (cdr lst)))))

I hope this isn't a homework
(defun idouble (x)
 (mapcar #'(lambda (n) (if (numberp n) (* 2 n) n)) x))

CL-USER 2 > (idouble nil)
NIL

CL-USER 3 > (idouble '(1 2 3))
(2 4 6)

CL-USER 4 > (idouble '(1 (2) 3))
(2 (2) 6)

CL-USER 5 > (idouble '(1 4 a e b 6))
(2 8 A E B 12)

bobihttp://tourdelisp.blogspot.com/

Wait the minute what do you mean by iterative solution?
(defun idouble (x)
(let (res)
(dolist (e x (nreverse res))
(push (if (numberp e) (* 2 e) e) res))))

bobi
.