CLISP recursive powers of cubes function that returns a list

Question

I'm looking for help in writing a function that returns a list of the first 15 cubes. Here's what I have so far, and I'm getting a stack overflow:

(defun cubes (dec lst) ;takes a number and a list as params
   (if (= dec 0) lst ;if dec is 0 return the list
       (cons (threes (- dec 1) lst))) ;;else decrement and call it again
)

I am calling it with the following code:

(cubes 15 nil) ;should print first 15 cubes

I just started on LISP today. Thanks for the help!

Don't you want to calculate cubes somewhere? – Svante Jan 27 '15 at 15:11 — Svante, Jan 27 '15 at 15:11

C. K. Young · Accepted Answer · 2015-01-27T15:26:17.783

2

Yeah, your function has a slight issue with it: :-)

You're recursing into threes instead of cubes.
You're trying to call cons with one argument (it requires two).
You're not changing the value of lst in the recursion, so, since the base case returns lst, you'll always get the initial lst value you passed in.

Here's a fixed version:

(defun cubes (dec lst)
  (if (zerop dec)
      lst
      (cubes (1- dec) (cons (* dec dec dec) lst))))

edited Jan 27 '15 at 15:26

answered Jan 27 '15 at 09:07

C. K. Young

219,335
46
382
435

1

Don't you want to calculate cubes somewhere? – Svante Jan 27 '15 at 15:12
@Svante Oops. Was too focused on getting the code working and neglected to look at the problem domain. – C. K. Young Jan 27 '15 at 15:24
1

@DomnWerner `(* dec dec dec)` is the same as `dec * dec * dec` in infix. It basically evaluates to the cube of `dec`. – C. K. Young Jan 27 '15 at 23:24

score 1 · Answer 2 · answered Jan 27 '15 at 13:53

1

You can also make use of the loop facility, and you SHOULD check if the initial value of dec is positive, otherwise you may end up with infinite loop/recursion:

(defun cubes (dec lst)
       (append
          (when (plusp dec)
                (loop for i from 1 to dec collect (expt i 3)))
          lst))

answered Jan 27 '15 at 13:53

tonso

1,760
1
11
19

You don't need `lst`, `append`, `when`. – Svante Jan 27 '15 at 15:13
I mean: `(defun cubes (n) (when (plusp n) (loop :for i :from 1 :to n :collect (expt i 3))))`. – Svante Jan 27 '15 at 16:36
That's right, but the author's function took an integer and a list, so suggested variants should as well. – tonso Jan 27 '15 at 17:50
`(defun cubes (n list) (declare (ignore list)) (when (plusp n) (loop :for i :from 1 :to n :collect (expt i 3))))` – Svante Jan 27 '15 at 18:35
1

The author's variant conses cubes to the list given as the second parameter, while yours just ignores it. When I wrote my variant, I took into account what the original variant actually did. – tonso Jan 27 '15 at 21:40
Chill out, guys. They both helped me learn a ton! Job well done, guys. – Domn Werner Jan 27 '15 at 23:26

CLISP recursive powers of cubes function that returns a list

2 Answers2