Common Lisp: cons inside loop

Question

I wonder why in the following code, d is not being consed into x. Any hints are much appreciated.

(defun it (x) 
  (setq f '(a b c))
  (dolist (d f)
    (cons d x))
  (print x))

Thank you!

Answer 1

I don't know a lot about LISP, but here's a few things I think I know that might help you:

• (cons d x) doesn't put d "into" x; it creates a new value that consists of d, followed by the contents of x. If x is a list, then (cons d x) will also be a list.

• (cons d x) doesn't change anything; it creates a value, but if you don't capture that value, it gets discarded again. In particular, that expression doesn't change x.

Answer 2

If I understand your intention properly, you want to append the list f comprising the symbols a, b, and c to the supplied argument x. How about this?

 (defun it (x)
   (append (if (listp x)
               x
               (list x))
           '(a b c)))

Note that cons doesn't modify any existing structures; it means "construct", and it creates fresh structure. In this case, you don't need to modify any anything, unless you're deliberately trying to write a destructive function that would modify the object pointed to by argument x in place. Such functions are rare and are usually named in a way to warn of such destructive behavior, such as nconc or nreverse. The leading n means "non-consing", suggesting it modifies existing structure rather than creating fresh structure.

Answer 3

You need to use PUSH to update a variable destructively. CONS just returns a new cons cell. If you don't do anything with that cons cell, then it is gone. Also not that you need to declare all variables. LET introduces a new local variable F, so LET is useful in your example. Just doing SETQ or SETF on an undeclared variable is not a good thing to do - actually in the Lisp standard, the consequences are undefined for such an operation.

(defun it (x) 
  (let ((f '(a b c)))
    (dolist (d f)
      (push d x))
    (print x)))

Or use a more functional approach.

(defun it (x)
  (print (append '(a b c) x)))

You also might want to throw in a call to REVERSE to get a different order of list elements.