Re: can s-exprs represent a graph without eval?

[Alan Bawden <address@hidden>]

   Date: Thu, 19 Jun 2003 16:07:25 -0400 (EDT)
   From: Guy Steele - Sun Microsystems Labs <Guy.Steele@sun.com>
   Cc: Guy.Steele@sun.com

   IIRC, Fred Blair, when he was working on IBM 370 Lisp back
   in the 1970s, actually put the analysis into the compiler
   to make it understand circular list structure in the code.
   So it would produce the same machine code for

   (prog (x)
	 (setq x 3)
      #1=(cond ((equal x 0) . #2#)
	       ((oddp x) (print x))
	       (t (setq x (- x 1)) . #1#))
	 (setq x (+ x 3))
      #2=(cond ((equal x 5) (return)))
	 . #1#))

   and

   (prog (x)
	 (setq x 3)
      A  (cond ((equal x 0) (go B))
	       ((oddp x) (print x))
	       (t (setq x (- x 1)) (go A)))
	 (setq x (+ x 3))
      B  (cond ((equal x 5) (return)))
	 (go A)))

   the point being the the first version would run a little faster
   when INTERPRETED.  Bleah.  De gustibus non disputandum.

   --Guy Steele

Surely that should be:

   (prog (x)
	 (setq x 3)
   . #1=((cond ((equal x 0) . #2#)
		((oddp x) (print x))
		(t (setq x (- x 1)) . #1#))
	 (setq x (+ x 3))
   . #2=((cond ((equal x 5) (return)))
	 . #1#))))

- Alan