letrec and set!

[Matthias Blume <blume@cs.Princeton.EDU>]

   Date: Thu, 19 Aug 93 18:05:10 +0200
   From: Christian Queinnec <Christian.Queinnec@inria.fr>
   Reply-To: Christian.Queinnec@inria.fr

      (letrec ((x (begin (set! y 1)
			 y ))
	       (y 3) )
	x )		 ---> 1

    But if letrec is operationally explained by a let with inner
   assignments, the way it is actually in r4rs, what is the reason to
   forbid writing to a mutually defined variable, nor it is necessary to 
   require a "smart" compiler.

No, the ``operational'' explanation is NOT the definition -- it only explains
the sematics of letrec as long as there is no violation of the rules.
One of these rules is the one that you are questioning.

   >    (letrec ((x (begin (assign-to-cell! y 1) (refer-to-cell y)))
   >             (y (make-cell 3)))
   >      x)

   Incidently I see the box model (or value cell) without explicit
   initialization so I see the program fragment more like:

   (let ((x (create-box))
	 (y (create-box)) )
     ; `let' to keep undeterminacy
     (let ((void1 (box-set! x (begin (box-set! y 1) (box-ref y))))
	   (void2 (box-set! y 3)) )
       (box-ref x) ) )

Indeed, this is a correct transformation.  But I claim that my
(shorter and more efficient) transformation would also be correct
PROVIDED that the rules are obeyed (which they are not in the example
program fragment).

-Matthias