Re: [Boost-users] shared_ptr invalid covariant return type

From: David Abrahams (dave_at_[hidden])
Date: 2008-06-24 21:19:22

John wrote:
> Thorsten Ottosen wrote:
>> John skrev:
>>> This pattern/idiom is well-established:
>>>
>>>   struct A {
>>>       virtual A*  clone () = 0;
>>>   };
>>>
>>>   struct B : public A {
>>>       virtual B*  clone ();
>>>   };
>>>
>>>
>>> But the above does not work when the pointers A* and B* are replaced
>>> by shared_ptr<A> and shared_ptr<B>:
>>>
>>>   struct  A {
>>>      virtual shared_ptr<A> clone () = 0;
>>>   };
>>>
>>>   struct  B : public A {
>>>      virtual shared_ptr<B> clone ();
>>>   };
>>>
>>>
>>> Is there a workaround? Another way to solve the same problem? Thanks!
>>
>> Well, don't use shared_ptr here. A newly created clone is not shared,
>> and it makes the interface less usable in other context to return a
>> shared ptr here.
>>
>> -Thorsten
> 
> 
> Hah! Well, yes, I guess it's true that there's always an option not to
> use smart pointers at all. :)  

auto_ptr would be more appropriate, and nearly as safe.

The other way to solve the same problem is to have the virtual clone
always return shared_ptr<A> and add another member function that returns
shared_ptr<B>.  Have the virtual function call that one.

Cheers,

-- 
Dave Abrahams
BoostPro Computing
http://www.boostpro.com