On 7/5/06, Thomas Lumley <tlumley at u.washington.edu> wrote:
On Wed, 5 Jul 2006, Gabor Grothendieck wrote:
On 7/5/06, Thomas Lumley <tlumley at u.washington.edu> wrote:
On Wed, 5 Jul 2006, Gabor Grothendieck wrote:
On 7/5/06, Thomas Lumley <tlumley at u.washington.edu> wrote:
On Tue, 4 Jul 2006, Gabor Grothendieck wrote:
In the code below, e is an environment which we copy to f and then
add attributes to e. Now f winds up with the same attributes.
In other words it seems that the attributes are a property of the
environment itself and not of the variable. Thus it appears we
cannot have two environment variables that correspond to the
original environment but with different attributes.
No, we can't. The two variables are references to the same environment,
they are the same.
If you want the attributes to be copies rather than references then
a list with the environment as an element and put the attributes on the
list.
I realize that that is how it works but what I was really wondering was
should it work that way?
I think it really should (and this question has come up before). If you
do
e<-environment()
f<-e
then there is only one object that f and e both point to. Now, since such
things as S3 class and matrix dimension are implemented as attributes I
think you really have to consider the attributes as part of the object
[which is also how they are implemented, of course]. So if e and f are
the same object they should have the same attributes.
I don't think this follows since in the other cases modifying the
object also creates a copy.
In cases other than environments, NULL, external pointers and weak
references a new object is (from the language definition point of view)
created on assignment. The fact that sometimes the actual memory
allocation is deferred is an implementation issue.
That is
e <- 2
f <- e
creates two different vectors of length 1, so of course they can have
different attributes.
For environments (and for NULL, external pointers, and weak references),
assignment does not create a new object. It creates another reference to
the same object. Since it is the same object, it is the same: attributes,
values, class, etc.
Another reasonable position would be to disallow attributes on
environments (as we do with NULL, another reference object), but that
seems extreme.
I don't think that that would solve it because there is still the issue
of the class attribute which you can't disallow.
Of course you can. It might be inconvenient, but it's not difficult.
In fact consider this:
e <- new.env()
f <- e
class(f) <- c("myenv", "environment")
F <- function(x) UseMethod("F")
F.environment <- function(x) 1
F.myenv <- function(x) 2
F(e) # 2
F(f) # 2
The point is that subclassing does not work properly with environments
No, subclassing *does* work. f and e are the same object, so they have the
same class, c("myenv", "environment"). The thing that doesn't "work" the
way you expect is assignment:
f <- e
doesn't create a new object, as it would for any other sort of object.
yet subclasses of the environment class should be possible since R
is supposed to be OO and I see no valid reason for exclusing environments
for this. I guess in this discussion I am coming to the realization that
this issue really is a problem with the current way R works.
It really is the way R is designed to work. Whether it is a problem or not
is a separate issue. Environments really are references, not values, and
they really work differently from the way most other objects work.
Yes, it about how you *try* to look at it and not how it was designed.
I was also bitten by this a few years ago. There are two ways to
think of 'e' and 'f' above:
1) 'e' and 'f' are R objects (reference variables) that refers to a
common object (the environment), and applying attr(e, name) <- value
applies to the R reference object 'e' and attr(f, name) <- value
applies to the 'f' object.
2) 'e' and 'f' are references to a common object (the environment)
and attr(e, name) <- value applies to the common environment object
and so do attr(f, name) <- value too.
Case 1) is wrong and Case 2) is correct.