Re: sharing list structure

[Denis Bueno <dbueno@gmail.com>]

On Thu, 24 Mar 2005 17:48:57 -0600, Joe Corneli
<jcorneli@math.utexas.edu> wrote:
> 
> I'm not sure how to do the following:

*snip*

> 
> So I guess what I want is an "implicit pointer" to A.
> 
> Looking at the box diagrams in the manual, it seemed to me that
> everything would be taken care of if I used "setcdr" to build the list
> B.  But that didn't quite work:
> 
> (progn
>   (setq A '(1 2 3))
>   (setq B (list 'foo))
>   (setcdr B A)
>   (setq A (append A (list 4)))
>   B)
> ;=> (foo 1 2 3)

The `append' doesn't alter the structure of the list A.

  (defvar *foo* (list 1 2 3))
  (append *foo* (list 4 5 6))
  *foo* => (1 2 3)

Hence, the result of append doesn't alter A's structure.

> If I handle A with kid gloves, then B comes out right:
> 
> (progn
>   (setq A '(1 2 3))
>   (setq B (list 'foo))
>   (setcdr B A)
>   (setcdr (nthcdr 2 A) (list 4))
>   B)
> ;=>(foo 1 2 3 4)

(setcdr (nthcdr ...) ...) _does_ alter A's structure. And thus it works.

> But is this the only way to go?  If it was possible, I would like to
> set things up so that I could do anything I wanted to do to A, and
> have B simply reflect that value at the end.

You can do "anything you want" with A, as long as any function you run
on A destructively modifies A. If it doesn't, then there's no way for
B to reflect the change.

So, just restrict yourself to destructive operations on A - like
setcdr, setcar, etc. - and you'll be set. Just note that A will always
have to be the "tail" part of B.

I.e. something like 

  B: (1 2 3 4 5 6)
  A: (3 4) ; a sublist of B

won't work, just because of the nature of lisp lists.

-- 
Denis Bueno
PGP: http://pgp.mit.edu:11371/pks/lookup?search=0xA1B51B4B&op=index