Hi Robert, Jeremy
Okay I have spotted the alien and the FBI. I was using 2.3.4 and
updated to 2.4 - how I could forget that the version of the OSG I was
using was something to investigate is beyond me.
The observer_ptr<> is now defined correctly and all my errors
automagically went away.
I am sorry for the noise.
Thank you Robert for the little lesson :)
Regards
Thibault
On Wed, May 14, 2008 at 5:19 PM, Robert Osfield
<[EMAIL PROTECTED]> wrote:
> Hi Thibault,
>
> On Wed, May 14, 2008 at 4:02 PM, Thibault Genessay <[EMAIL PROTECTED]> wrote:
>> I really don't know why it is this way - I had actually never noticed
>> that ref_ptr<> had this "feature". It makes more sense to me how the
>> observer_ptr<> is implemented because ref_ptr and observer_ptr are
>> just proxies that should IMHO behave like ordinary references or
>> pointers when we call operator*() or operator->(). (I'm sure a C++
>> guru could slap me for what I've said, but I'd enjoy it if it came
>> with an explanation :)
>
> Which version of the OSG are you using? In 2.4 the observer_ptr<> is
> consistent with
> ref_ptr<> i.e.
>
> inline T& operator*() const { return *_ptr; }
> inline T* operator->() const { return _ptr; }
> inline T* get() const { return _ptr; }
>
> The next question is this wacky world is why does ref_ptr<> and
> observer_ptr<> do this trick?
>
> Try:
>
> typedef std::set< osg::ref_ptr<Node> > MySet;
>
> Then try to access members of it, and you'll find out they are in fact
> now a const ref_ptr<> and you
> can't access any non const methods of Node. What the const method
> here is const of the ref_ptr<>
> not const of what its pointing to. If you meant this then you'd have
> osg::ref_ptr<const Node>.
>
> However, it all can get a bit sticky, as sometimes you intend const
> osg::ref_ptr<Node> to be const osg::ref_ptr<const Node> in which case
> the relaxation of the the above accessors misses this.
>
> Robert.
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>
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