Re: [boost] Smart Pointer new/Forwarding Problem Solved (by Force)

From: Thorsten Ottosen (thorsten.ottosen_at_[hidden])
Date: 2006-03-29 16:28:44

Fernando Cacciola wrote:
> David Abrahams wrote:
> 
>>David Abrahams <dave_at_[hidden]> writes:
>>
>>
>>>Ulrich Eckhardt <doomster_at_[hidden]> writes:
>>>
>>>
>>>>On Wednesday 29 March 2006 08:05, Václav Veselý wrote:
>>>>
>>>>>I'm confused with syntax. new_<T> always creates auto_ptr<T>. How
>>>>>can I create for example shared_ptr<T>?
>>>>
>>>>You probably can't, but:
>>>> - std::auto_ptr is much less resource intensive (shared_ptr
>>>>requires an additionally, dynamically-allocated structure to hold
>>>> some internals) - you don't need to, as auto_ptr converts to
>>>>shared_ptr, there is a special ctor taking an auto_ptr
>>>
>>>Unfortunately for this particular facility, the converting
>>>constructor
>>>is explicit, so you can't do
>>>
>>>   f(new_<T>(a, b, c))
>>>
>>>if f takes a shared_ptr<T>.
>>
> Then isn't it worth generalizing new_<> to all sorts of smart pointer and 
> handles?
> 
> Like:
> 
>  f( new_< any_smart_ptr<T> >(a, b, c)) ;

this might be worth persuing (see below)

> As long as there's a way to get the 'element_type' from 'any_smart_ptr' 
> it'll work.
> 
> On top of that there could be the friendlier: make_auto_ptr, 
> make_shared_ptr, etc...

That would generate *a lot* of code to parse for the compiler.

It is better that new_<T>(...) produces an unspecified type
that has a templated conversion operator, eg.:

template< class T >
struct new_return
{
    std::auto_ptr<T> new_;

    new_return( T* new_ ) : new_(new_)
    { }

    template< class SP >
    operator SP() const
    {
      return SP( new_.release() );
    }

    operator std::auto_ptr<T>() const
    {
      return new_;
    }
};

Or something

-Thorsten