[REBOL] Re: Does REBOL cons?
From: massung:g:mail at: 15-Mar-2006 16:48
Hehe. Let me explain why I'm a bit confused and maybe that will help. I just
want to build off what you originally replied with.
Okay, so, the definition of a cons, of course, is just a trivial,
uni-directional, linked-list:
(A . (B . (C . NIL))) -> (A B C)
(D . NIL) -> (D)
The reason that the original list in FOO changes (assuming that REBOL uses
conses) is that the cons holding 'C' is modified to also include the D cons:
(A . (B . (C . (D . NIL)))) -> (A B C D)
And, of course, since the function FOO just makes use of that pre-allocated
list in bytecode (or whatever REBOL uses for interpretation of data), the
source
of FOO changes with each call.
However, [d] is no different from [a b c] in the source (meaning that both
are "pre-allocated data" - for lack of a better term - they aren't recreated
each call). Suppose I wrote the function like this:
foo: func [/local a b] [
a: [a b c]
b: [d]
append a b
]
Now, after the first call, the function's source is now:
foo: func [/local a b] [
a: [a b c d]
b: [d]
append a b
]
Simple enough. But, the last cons in the list that 'a' points to is the same
cons that 'b' points to (if REBOL is consing). So, after the second call, we
should have an infinite list:
foo: func [/local a b] [
a: [a b c d d d d d ... ]
b: [d d d d d d ... ]
append a b
]
This is what Lisp does, and if REBOL conses, is what I would expect that it
would do as well.It could also be that APPEND has a built-in safe guard, in
that maybe it copies the second argument to prevent self-referencing
objects?
Anyway, perhaps this better explains my question. :-)
Jeff M.
--
massung-gmail.com
On 3/15/06, Izkata <Izkata-comcast.net> wrote:
