Re: (iota N) in Spice Lisp idiom

On 2024-05-26, B. Pym <No_spamming@noWhere_7073.org> wrote:

On 2/27/2024, Kaz Kylheku wrote:
>

Common Lisp's loop macro has a collect{ing} clause.
 
  (defun iota (n)
    (loop for i from i to n
          collect i))

>
>
Incorrect; three errors.

I only see one: it was intended to be 1 to n, not i to n.  With that
correction it works to the extent that (iota 3) produces (1 2 3).

Correct:
>
(defun iota (n)
  (loop for i from 0 below n
    collect i))

In that post, I decided that the requirements for (iota n)
are to go from 1 to n. That's what is correct, as far as the
code goes. In my post, deciding what iota means is my privilege.

While you've gone to fuck yourself, remember to practice your
counting.