Re: [boost] [thread] condition suffering from lost wakup bug?

From: Matt Hurd (matt.hurd_at_[hidden])
Date: 2005-09-14 19:43:39

> On 15/09/05, Roland Schwarz <roland.schwarz_at_[hidden]> wrote:

> Thank you,
> Roland
> 
> #include <iostream>
> 
> #include <boost/thread.hpp>
> 
> using namespace boost;
> 
> mutex guard;
> typedef mutex::scoped_lock lock_type;
> condition cond;
> 
> long global_count = 0;
> long count_for_test = 10000;
> 
> long global_number_of_threads = 0;
> long number_of_threads_for_test = 1000;
> 
> bool global_flag = true;
> 
> void do_work()
> {
>     lock_type lock(guard);
>     ++global_number_of_threads;
>     cond.notify_one();
>     while (global_count < count_for_test)
>         cond.wait(lock);
> }
> 
> void do_flag()
> {
>     lock_type lock(guard);
>     ++global_number_of_threads;
>     cond.notify_one();
>     while (global_count < count_for_test) {
>         cond.wait(lock);
>         global_flag = false;
>         // Step 3: The following signal occasionally
>         // is beeing lost. See below.
>         cond.notify_one();
>     }
> }
> 
> int main()
> {
> 
>     thread_group pool;
> 
>     for (long i=0; i < number_of_threads_for_test; ++i)
>     {
>         pool.create_thread( do_work );
>     }
>     pool.create_thread( do_flag );
> 
>     { // we wait until all threads are up
>         lock_type lock(guard);
>         while (global_number_of_threads <
>             number_of_threads_for_test+1)
>             cond.wait(lock);
>     }

I can't see why you get here?  I might be missing something but you're
likely to end up with all threads waiting on conditions here...  All
the do_work threads are waiting, the do_flag is waiting.  Then if all
the threads aren't up and running we, the main thread, wait on the
condition too, but we might not be the one the consumes a notify_one
as the others threads start so we never progress.  I could have the
logic arse about though...

matt.