Replace elements in nested quoted lists adds new elements?

Question

I have a nested list, and I am trying to non-destructively replace all its elements (inside the nested list as well). That is, given my input list

\'(1 \'(2         


        

Answer 1

You have to realize that 'foo is syntactic sugar for (quote foo), so when you use quotes inside an already-quoted list, this:

'(1 '(2 3 4) '(5 6 7) 8 9)

evaluates to this:

(1 (quote (2 3 4)) (quote (5 6 7)) 8 9)

So when you substitute all the list elements with 0, you get:

(0 (0     (0 0 0)) (0     (0 0 0)) 0 0)

You either need to not put extra quotes in your examples, or you need to handle the quote operator specially in subs-list:

(defun subs-list (list value)
  "Replaces all elements of a list of list with given value"

  (loop for elt in list
       collect
         (cond 
           ((listp elt) 
             (subs-list elt value))
           ((eq 'quote elt)
             elt)
           (t
             value))))