Thanks a lot Wacek for this clear description of the problem - I was
not aware, that it is that complex.
I definitely did not consider the initialize() function in writing my code.
But as I only want to allocate the space for the objects, it does not
matter here. But when I write a simulation and want to initialize it
with two hundred different individuals, this definitely becomes
crucial - thanks again for this clarification and the very useful
references (especially [2] for the moment).
Am I right that all assignments of classes via <- or -> are by value
and NOT by reference? So when
setClass('foo',
representation=representation(content='character'),
prototype=list(content='foo'))
A <- new("foo")
B <- A
B is a different object then A, i.e. changes in B are not reflected in
A and vice versa, but they have the same content.
It seems that if foo has a slot O containing an object of class
"fooInFoo", this object O also copied by value. I.e. when, following
above,
setClass('foo2',
representation=representation(content='character'),
prototype=list(content='fooInFoo'))
setClass('foo',
representation=representation(content='character', O='foo2'),
prototype=list(content='foo', O=new('foo2')))
A <- new("foo")
B <- A
a...@o@content <- "AAAAAAAAAAAAAA"
a...@o@content
[1] "AAAAAAAAAAAAAA"
b...@o@content
[1] "fooInFoo"
Is this always the case? and is there a way of copying by reference
(or passing an object to a function by reference)?
Rainer
Cheers
Rainer
On Mon, Jun 1, 2009 at 11:36 PM, Wacek Kusnierczyk
<waclaw.marcin.kusnierc...@idi.ntnu.no> wrote:
> Wacek Kusnierczyk wrote:
>>
>> consider:
>>
>> setClass('foo',
>> representation=representation(content='environment'))
>> setMethod('initialize', 'foo', function(.Object) {
>> .obj...@content = new.env()
>> .obj...@content$name = 'foo'
>> .Object })
>>
>> foos = rep(list(new('foo')), 2)
>> foos[[...@content$name = 'bar'
>> foos[[...@content$name
>> # 'bar'
>>
>> of course, because you have just one object twice on the list, its
>> content is an environment, and with environments r does not pretend to
>> be functional. but:
>>
>> foos = lapply(1:2, function(x) new('foo'))
>> foos[[...@content$name = 'bar'
>> foos[[...@content$name
>> # 'foo'
>>
>> of course, because you have two distinct objects, and assignment to one
>> of them has no effect on the other. similarly if 'foo' is defined as
>> follows:
>>
>> setClass('foo',
>> representation=representation(content='environment'))
>> setMethod('initialize', 'foo', function(.Object) {
>> .obj...@content$name = 'foo'
>> .Object })
>>
>> or as follows:
>>
>> setClass('foo',
>> representation=representation(content='environment'),
>> prototype=list(content={
>> e=new.env()
>> e$name='foo'
>> e }))
>>
>>
>
> actually, not the last one; i got caught by that redefining 'foo' with
> setClass does not remove the initializer. so here's another situation
> (best to execute in a fresh session) you should beware:
>
> setClass('foo',
> representation=representation(content='environment'),
> prototype=list(content={
> e=new.env()
> e$name='foo'
> e }))
>
> foos = lapply(1:2, function(x) new('foo'))
> foos[[...@content$name = 'bar'
> foos[[...@content$name
> # "bar"
>
> in this case, even if you're careful enough to create two objects, they
> share the representation because they both get it from the same prototype.
>
> vQ
>
>
>
--
Rainer M. Krug, Centre of Excellence for Invasion Biology,
Stellenbosch University, South Africa
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