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From: Ovanes Markarian (om_boost_at_[hidden])
Date: 2006-08-24 11:23:09
There are many things which boost::threads do not have (event variables, some mutex types leading
to deadlocks were removed). As I read in the docs, the main idea was to design _robust_ threads.
Performance also plays a huge role. If you look into the implementation of thread::join you will
see that there is no mutex defined. So it is not safe to access join() from multiple threads,
since one thread could wait on object which was finished anyway:
void thread::join()
{
assert(m_joinable); //See race condition comment below
int res = 0;
#if defined(BOOST_HAS_WINTHREADS)
res = WaitForSingleObject(reinterpret_cast<HANDLE>(m_thread), INFINITE);
assert(res == WAIT_OBJECT_0);
res = CloseHandle(reinterpret_cast<HANDLE>(m_thread));
assert(res);
#elif defined(BOOST_HAS_PTHREADS)
res = pthread_join(m_thread, 0);
assert(res == 0);
#elif defined(BOOST_HAS_MPTASKS)
OSStatus lStatus = threads::mac::detail::safe_wait_on_queue(
m_pJoinQueueID, NULL, NULL, NULL, kDurationForever);
assert(lStatus == noErr);
#endif
// This isn't a race condition since any race that could occur would
// have us in undefined behavior territory any way.
m_joinable = false;
}
With Kind Regards,
Ovanes Markarian
On Thu, August 24, 2006 16:03, Aubrey, Jason wrote:
> I can accept that I should use a condition variable for this situation
> but I'm still not understanding the reason why.
>
> If the intent of join is to block until the thread has completed, why
> doesn't it just return when it realizes the thread has completed? Also
> how is the second call any different from the first call occurring after
> the thread has completed?
>
> Apparently the C# designers had my same perspective since the following
> completes without error:
>
> // C# sample program to demonstrate join() semantics
> using System.Threading;
> namespace Test1
> {
> class Sample
> {
> void f(){}
> static void Main(string[] args)
> {
> Sample s = new Sample();
> Thread t = new Thread(new ThreadStart(s.f));
> t.Start();
> t.Join();
> t.Join(); // No exception is thrown here, call simply
> returns
> }
> }
> }
>
> Regards,
> Jason
>
> -----Original Message-----
> From: boost-users-bounces_at_[hidden]
> [mailto:boost-users-bounces_at_[hidden]] On Behalf Of Ovanes
> Markarian
> Sent: Thursday, August 24, 2006 4:25 AM
> To: boost-users_at_[hidden]
> Subject: Re: [Boost-users] [Boost-Threads] Assertion during a second
> call to thread::join()
>
> I think what you need is a condition variable and not join.
> You can use notify_one and notify_all member of a common condition
> variable to wake up one or respective all waiting threads:
>
> http://www.boost.org/doc/html/condition.html
>
>
> Please read this article in DDJ about boost threads there is written how
> to handle different threading issues:
> http://www.ddj.com/dept/cpp/184401518
>
>
> With Kind Regards,
>
> Ovanes Markarian
>
>
> On Thu, August 24, 2006 09:05, Sascha Seewald wrote:
>>
>>
>> Aubrey, Jason wrote:
>>> Hi,
>>>
>>> Does anyone know the reason for the behavior in the following example
>
>>> program?
>>>
>>> This example attempts to simulate the scenario where two threads wait
>
>>> on a third thread. However, I realize that the following is actually
>
>>> one thread waiting on a second thread twice.
>>>
>>> Regards,
>>> Jason
>>>
>>> My Environment: MSVC 8.0, boost v1.33.1
>>>
>>> #include <boost/thread/thread.hpp>
>>> #include <boost/thread/xtime.hpp>
>>>
>>> void f()
>>> {
>>> }
>>>
>>> void main()
>>> {
>>> boost::thread t(f);
>>> t.join();
>>> t.join(); // Causes an assertion
>>
>> A call to thread::join() requires the thread to be joinable. After the
>
>> first call to join() the thread has finished and is not joinable
> anymore.
>> If you'd like to execute f twice, create a seperate thread for each
> run.
>>
>> hth
>>
>> Sascha
>>
>> _______________________________________________
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>> Boost-users_at_[hidden]
>> http://lists.boost.org/mailman/listinfo.cgi/boost-users
>>
>
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