Re: [Boost-users] shared_ptr and covariant return types

From: Ovanes Markarian (om_boost_at_[hidden])
Date: 2007-11-08 09:48:52

Hi!

do_clone is really private ;) Your derived class has public clone member and
you should use it from the free function, but no the do_clone.

Regards,
Ovanes

On 8 Nov 2007 14:53:49 +0100, Bernhard Wagner boost <
lists.boost.org_at_[hidden]> wrote:
>
> In this post Peter Dimov shows a solution for the covariant return type
> problem when using shared_ptr:
> http://lists.boost.org/boost-users/2003/02/2996.php
>
> I'm interested in the free function solution he provides:
>
> > [snip] you can use the usual pattern from the no-covariant-return days:
> >
> > class B
> > {
> >     virtual shared_ptr<B> do_clone() const = 0;
> > };
> >
> > class X: public B
> > {
> >     virtual shared_ptr<B> do_clone() const { return shared_ptr<X>(new
> > X(*this)); }
> >
> > public:
> >
> >     shared_ptr<X> clone() const
> >     {
> >         shared_ptr<X> px = dynamic_pointer_cast<X>(this->do_clone());
> >         assert(px);
> >         return px;
> >     }
> > };
> >
> > It buys you some extra safety, too.
> >
> > You could also use a free function:
> >
> > template<class X> shared_ptr<X> clone(X const & x)
> > {
> >
> >     shared_ptr<X> px = dynamic_pointer_cast<X>(x.do_clone());
> >     assert(px);
> >     return px;
> > }
> >
>
> So, to clone an object using the clone method, I could use:
>
> X* x;
> shared_ptr<X> spX(x = new X());
> shared_ptr<X> spX2 = spX->clone();
> // or:
> shared_ptr<X> spX3 = x->clone();
>
> But how is the free function meant to be used?
>
> This provokes the compiler error: '... do_clone() const is private ...':
> shared_ptr<X> spX4 = clone(*spX);
>
> Thank you
> Bernhard
>
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