Re: [boost] [interest] rich-typed smart pointers

6 Oct 2013

      On 10/05/13 16:26, Julian Gonggrijp wrote:
...
Larry Evans wrote:
...
...
https://github.com/jgonggrijp/rich-typed-pointers
[snip]
If there were a templated owned_ptr CTOR:
template<typename Derived>
    owner_ptr (owner_ptr<Derived> && source) : pointer(source.pointer) {
        source.pointer = nullptr;
    }
wouldn't that eliminate the need for the make_dynamic templated function?
Just wondering.  I haven't tried it, but I'd guess it would work because
the ownd_ptr<Derived> could be constructed with the:
make<Derived>(t1,t2,...tn)
and then the templated owned_ptr CTOR would then convert that to the
base class?
I agree this would be great, but unfortunately it appears it can't be done.
owner_ptr<Derived> would have to friend-declare owner_ptr<Base>,
I don't see why.  AFAICT, you'd require *fewer* friend declarations, not 
more( you wouldn't need the:
template 
     friend owner_ptr<U1> make_dynamic (Us&& ...);

the only friend needed would the existing:

     template 
     friend owner_ptr<U> make (Us&& ...);

that's because the proposed templated constructor would have the same
access as the existing:

     owner_ptr (owner_ptr && source);

CTOR, i.e. public.  IOW, with the proposed templated CTOR, the
following example code creating an owner_ptr<Base> should work,
AFAICT:

     owner_ptr<Base>
   owner_baee
     ( make<Derived>
       ( DerivedInit0()
       , DerivedInit1()
       .
       .
       . DerivedInitN()
       )
     );

Where DeriveInit0, DerivedInit1, ... DerivedInitN
are the types of the args to the CTOR for Derived.

Is there some reason why the above example code would
*not* compile, given then proposed templated CTOR?
[snip]

-regards,
Larry