[plt-scheme] Question about variable scope in lists

From: John Clements (clements at brinckerhoff.org)
Date: Sat Feb 26 10:20:57 EST 2005

On Feb 26, 2005, at 12:33 AM, A00453721 at itesm.mx wrote:

>   For list-related administrative tasks:
>   http://list.cs.brown.edu/mailman/listinfo/plt-scheme
> Greetings. First of all, I would like to clarify that, while I am a CS 
> student,
> I'm not explicitly asking for help on my assignment. I would just like 
> an
> explanation of why is this happening:
> Currently, I'm investigating the effects of destructive modifications 
> on
> variables passed to procedures using Dr. Scheme.
> While reading the "Teach yourself Scheme in fixnum days" study book, I 
> found
> this destructive code which reverses a list "in-place".
> (define reverse!
>   (lambda (s)
>     (let loop ((s s) (r '()))
>       (if (null? s) r
>       (let ((d (cdr s)))
>             (set-cdr! s r)
>         (loop d s))))))
> However, I tried it in the interactions windows like this:
> ( define U '( 1 2 3 4 5 6 7 8 ) )
> ( reverse! U )
> Not to my surprise, the output was:
> (8 7 6 5 4 3 2 1)
> but when I asked the interpreter to reevaluate U, it just displayed:
> (1)

It's nice to keep an understanding of the difference between changing a 
binding and changing a value.  The first one changes what value an 
identifier is bound to, the second changes the value itself.

Nowhere in this program do you do a "(set! U ...)" or a second "(define 
U ...)", so you're not changing the binding for U.  U still refers to 
the same value that it did at the beginning of the computation.

But wait!  At the beginning, U was bound to `(1 2 3 4 5 6 7 8), and now 
it's bound to `(1)... how can that be the "same value"?  Well, in a 
world with set-car! and set-cdr!, you have to realize that values can 
change.  So, U was bound to a cons pair whose first element was 1 and 
whose second element was another value that represents `(2 3 4 5 6 7 
8).  After running the in-place reverse, you've done a set-cdr! on this 
cons pair, so that its second element is now the null value.

Draw some pictures to convince yourself, if it's not already clear.  
This would be a great use for Slideshow.


p.s.: if you want to see U refer to the new value, you could just do it 
like this:  (set! U (reverse! U))

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