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[REBOL] Does REBOL cons?

From: massung::gmail at: 15-Mar-2006 16:15

Okay, so, I've been trying something out, comparing REBOL to Lisp. The functions in REBOL make it appear as though the lists (series!) in REBOL are just your average conses (first, next, associations, etc). Because of this and constant data optimizations, etc, this function in REBOL makes sense: foo: func [/local a] [ a: [a b c] append a [d] ]
>> foo
== [a b c d]
>> foo
== [a b c d d] This makes sense, at least, until after the second call. The above function would be akin to the following Lisp code, which does *almost* the same thing: (defun foo () (let ((a '(a b c))) (nconc a '(d)))) The main difference being that the second call to (foo) will result in an infinite list, because '(d) is scoped data, just like '(a b c), and the second call will cons '(d) with itself, giving us '(a b c d d d d d ...). In REBOL, everything seems to match up, except that [d] doesn't cons with itself. Now, IMO, this is good, desired behavior (not generating an infinite list). However, what this implies "under-the-hood" is that REBOL doesn't use traditional consing for lists, but does something else instead. I was wondering if anyone could shed a little light on this. If REBOL does use consing, then why don't I end up with an infinite, self-referencing list? If it doesn't using consing, what does it use (not that it matters, I'm just curious)? This is actually a little exciting to me, because if it isn't using your typical consing, then that means REBOL's associated lists could potentially be a lot faster than Lisp's (which, of course, have O(n) access times). Jeff M. --