Common Lisp: is delete-if the same as setf + remove-if?
common-lisp, lisp, primes
Solution
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.
Problem
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?