Re: "push" creating circular objects

[Pascal J. Bourguignon]

Charles Sebold <csebold@gmail.com> writes:

> On Aug 20, 4:50 pm, weber <hug...@gmail.com> wrote:
>
>> Minor variation:
>>
>> (defun test2 (str)
>>   (let (my-list)
>>         (with-temp-buffer
>>           (insert str)
>>           (goto-char (point-min))
>>           (while (not (eobp))
>>                 (cond
>>                   ((= (char-after) ?#)
>>                    (push "ol" my-list))
>>                   ((= (char-after) ?*)
>>                    (push "ul" my-list)))
>>                 (forward-char 1)))
>>         my-list))
>>
>> no need to bind my-list to nil too :)
>
> Useful tip, thanks.
>
> Astoundingly I'm still having the same problem, when I run this in the
> context of the larger program.  Against the string "** " it returns
> the circular list (#1="ul" #1#).  In the scratch buffer it works (same
> as with my function earlier).

This is not a circular list, so whatever you may try or infer from
that will be wrong.


> So, I must be rebinding something else in a bad way, but heck if I
> know how.  I'm going to try to trim the whole problem down to a bare
> minimum test case and then if I can't figure it out I'll post the
> whole thing and ask people to eval and see what they can come up
> with.  Thanks for your time.

(defun circular-length (list)
  "LIST must be either a proper-list or a circular-list, not a dotted-list.
RETURN: the total length ; the length of the stem ; the length of the circle.
"
  (let ((indexes (make-hash-table)))
    (loop
       for i from 0
       for current on list
       do (let ((index (gethash current indexes)))
             (if index
                 ;; found loop
                 (return (values i index (- i index)))
                 (setf (gethash current indexes) i)))
       finally (return (values i)))))


(circular-length '(a b c d e))
--> (5)

(circular-length '(#1=a b #1# d #1#))
--> (5)

(circular-length '#1=(a b c d e . #1#))
--> (5 0 5)

(circular-length '(a b . #1=(c d e . #1#)))
--> (5 2 3)


Since (#1="ul" #1#) is not a circular list, you can print it without 
print-circle:

(defun print-safely (object)
   (let ((print-circle  (rest (circular-length object))))
      (print object)))

(print-safely '(a b c d e))
prints: (a b c d e)
       
(print-safely  '#1=(a b c d e . #1#))
prints: #1=(a b c d e . #1#)

(print-safely '(#1=a b #1# d #1#))
prints: (a b a d a)

(print-safely '(#1="ul" #1#))
prints: ("ul" "ul")


So what's the difference between ("ul" "ul") and (#1="ul" #1#)?

It's that in ("ul" "ul") we have two different string objects that
happen to contain the same characters in the same order, while in
(#1="ul" #1#) we have the same string object twice:

(let ((items '(#1="ul" #1#)))
   (eq (first items) (second items)))
--> t

(let ((items '("ul" "ul")))
   (eq (first items) (second items)))
--> nil



-- 
__Pascal Bourguignon__                     http://www.informatimago.com/

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