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From: George A. Heintzelman (georgeh_at_[hidden])
Date: 2001-05-19 12:13:52

> --- In boost_at_y..., "Rainer Deyke" <root_at_r...> wrote:
> > The correct value if 'bp' at this point is '0'. Your function does
> not
> > mirror this behavior. The following fixes that problem:
> >
> > template<typename T, typename U>
> > inline shared_ptr<T>& sp_dynamic_cast(const shared_ptr<U>& a)
> > {
> > if (dynamic_cast<T *>(a.get()) {
> > return *(shared_ptr<T>*)&a;
> > }
> > return shared_ptr<T>(0);
> > }
> Agreed about the returning 0. But isn't there a problem here with
> returning a temporary? I changed ito to inline shared_ptr<T>&
> sp_dynamic_cast(const shared_ptr<U>& a)

I don't believe that this code is correct. You are assuming that the
pointer to U included internally inside the shared_ptr a is also valid
as a pointer to T, by forcing the compiler to assume that a can be
treated as a shared_ptr to T as well as to U. But this may not be the
case, particularly in the presence of multiple inheritance (where
compilers often have to do pointer arithmetic to handle pointer casts).
Because of this, I don't think you are going to be able to return a
reference to a shared_ptr from this kind of function; you should go
through the system's dynamic_cast mechanism and return a shared_ptr
based on that result. This will I think require some mechanism in the
shared_ptr for construction using another shared_ptr's reference count.
How to do this safely needs some figuring out.

I note that I would welcome having dynamic_cast available for smart
pointers (and don't neglect std::auto_ptr when you do it), so I think
the effort to do this is good, but I think that the attempt here will
fail as it stands.

George Heintzelman

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