How to reverse a list such that every sublist is also reversed? This is what I have so far:
(defun REV (L)
(cond
((null L) nil)
((listp L)
(append
(REV (cdr L))
(list (car L))))
(t
(append
(REV (cdr L))
(list (car L))))))
You are on the right track, but your last two conditions have the same action, which should give an indication that one of them is not doing what it should. Indeed, the second condition, the listp case, is not right, because when it's a list, you need to append the reverse of that list instead of the unmodified list. A possible solution:
(defun reverse (l)
(cond ((null? l) nil)
((listp (car l)) (append (reverse (cdr l))
(list (reverse (car l)))))
(t
(append (reverse (cdr l))
(list (car l))))))
> (reverse '((1 2 3) (4 5 6)))
((6 5 4) (3 2 1))
As you can see, the only difference is that you test if the first element is a list, and if it is, you reverse the first element before appending it.
I'd write it this way:
(defun reverse-all (list)
(loop
with result = nil
for element in list
if (listp element)
do (push (reverse-all element) result)
else do (push element result)
finally (return result)))
Sounds like a homework problem :)
Looks like you started by writing the regular reverse code. I'll give you a hint: The second condition (listp L) isn't quite right (it'll always be true). You want to be checking if something else is a list.
dmitry_vk's answer (which probably is faster in most lisps than using append in the previous examples) in a more lispish way:
(defun reverse-all (list)
(let ((result nil))
(dolist (element list result)
(if (listp element)
(push (reverse-all element) result)
(push element result)))))
Or even:
(defun reverse-all (list)
(let ((result nil))
(dolist (element list result)
(push
(if (listp element) (reverse-all element) element)
result))))
