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From: Ovanes Markarian (om_boost_at_[hidden])
Date: 2008-01-16 02:31:16
Try using the enable_shared_from_this idiom:
http://www.boost.org/libs/smart_ptr/sp_techniques.html#from_this
This will allow you to pass a raw pointer to your thread and later on bind
the raw pointer to the correct shared counter.
Good Luck,
Ovanes
On Jan 16, 2008 6:45 AM, Elli Barasch <comptonsw_at_[hidden]> wrote:
> Richard Dingwall wrote:
>
> On Jan 16, 2008 5:22 PM, Elli Barasch <comptonsw_at_[hidden]> <comptonsw_at_[hidden]> wrote:
>
>
> Richard Dingwall wrote:
>
> On Jan 16, 2008 1:11 PM, Elli Barasch <comptonsw_at_[hidden]> <comptonsw_at_[hidden]> wrote:
>
>
> I'd like to pass a shared_ptr as the entry argument to a thread function
> via pthread_create. However,
> the entry argument is prototyped as void *. I can't simply cast the
> shared pointer as a (void *). How do I go about this?
>
> Try passing:
>
> static_cast<void *>(your_ptr.get())
>
>
> Richard
> _______________________________________________
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>
>
> A follow up question. Does get() increase the reference count, or does it
> simply return the underlying pointer without doing so? If so, I'd be
> worried that a race exists such that the object could be destroyed before
> the thread gets to run.
>
>
> get() does not incremement the reference count, as it returns a raw
> pointer, and there is no way of knowing how a raw pointer is being
> consumed or stored.
>
> If you the function you are passing it to retains a reference to the
> pointer anywhere, there is no way this can be tracked by the
> shared_ptr. Furthermore, if the shared_ptr(s) go out of scope the raw
> pointer references will be left dangling.
>
> HTH,
>
> Richard
> _______________________________________________
> Boost-users mailing listBoost-users_at_[hidden]http://lists.boost.org/mailman/listinfo.cgi/boost-users
>
> Is there a way I can "new" a copy of the shared_ptr onto the stack?
> Something like:
>
> pthread_create(&t,&attr,&func,(void *) new shared_ptr<T>(shp));
> //mangled syntax, I know
>
> Something like this would guarantee the object stays in scope?
>
>
>
>
>
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