Question

The following code generates prime from 1 to n:

(defun prime-list(n)
  (let ((a)(b)(x (floor (sqrt n))))
    (loop for i from (floor n 6) downto 1 do
          (push (1+ (* 6 i)) a)
          (push (1- (* 6 i)) a))
    (loop while (<= (car a) x) do
          (push (car a) b)
          (setf a (remove-if #'(lambda(m)(or (= 0 (mod m (car a))) (> m n))) a)))
    (append '(2 3) (reverse b) a)))

It seems to me the part

(setf a (remove-if #'XXX a)) 

can be replaced by

(delete-if #'XXX a)

And I hoped this would make it faster. However when I made that change the function now get into an infinite loop and never returns. Why?

Answer 1

As mentioned in the comments, you need to set the variable.

DELETE-IF is mostly a destructive version of REMOVE-IF. REMOVE-IF returns a freshly consed sequence, which does not contain the removed elements. DELETE-IF may return a sequence which is reused.

If you have a variable, which is bound to a list, you still need to set the result. Above functions return results, but they don’t set variables to the result. In case of a list, the result of a DELETE-IF operation can be the empty list and there is no way the side effect can be, that a variable can be set to it – when it was pointing to a non-empty list.