[boost] Re: how do I release shared_ptr ?

From: Joe Gottman (jgottman_at_[hidden])
Date: 2004-01-07 23:12:13

"David Abrahams" <dave_at_[hidden]> wrote in message
news:ur7yb6t0m.fsf_at_boost-consulting.com...
> "Joe Gottman" <jgottman_at_[hidden]> writes:
>
> > "David Abrahams" <dave_at_[hidden]> wrote in message
> > news:u8ykjepph.fsf_at_boost-consulting.com...
> >> "Peter Dimov" <pdimov_at_[hidden]> writes:
> >>
> >> > Jason House wrote:
> >> >> Peter Dimov wrote:
> >> >>
> >> >>> When multiple threads are involved there is a race condition
between
> >> >>> the unique() test and the actual release, and I'm not sure how such
> >> >>> a feature would interact with a lock-free implementation.
> >> >>
> >> >>
> >> >> I think this statement might be incorrect.  If nothing else, I think
> >> >> some careful thought about it might yield an easy solution to the
> >> >> problem. If I understand right, a shared_ptr will be unique if and
only
> > if
> >> > it
> >> >> is the only shared_ptr to that object.  In that scenario, there will
be
> >> >> no other shared_ptr's in any other threads that could cause the
> >> >> reference count to change from 1 at any instant in time...
> >> >
> >> > You're almost right, except for one small detail; weak pointers do
not
> >> > affect unique(), and can generate other shared_ptr copies.
> >>
> >> Maybe we want weak_unique(), or we want to change unique() so that it
> >> returns weak_count.
> >>
> >> We should look again at the use cases for unique().  Maybe they're
> >> supported just as well or better by weak_count.
> >>
> >
> >     One major use for unique() is copy-on-write.  The code would look
> > something like:
> >
> >     void write_to(shared_ptr<Foo> p, const Foo &newValue)
> >     { // For simplicity, assume p.get() != 0
> >         if (p.unique()) {
> >             *p = newValue;
> >         } else {
> >             p.reset(new Foo(newValue));
> >         }
> >     }
> >
> > Under the current definition of unique(), this code is impossible to
make
> > thread-safe if weak_ptrs might exist.
>
> Would it work just as well if unique() returned weak_count?
>

   It would work better. If nobody is using weak_ptrs then weak_count equals
1 if and only if shared_count equals 1.  And if weak_ptrs are possible, then
having unique() depend on weak_count guarantees that this code is
thread-safe.

Joe Gottman