Re: Re: Newbie needs pointer to currently executing thread.

From: Dave Moore (yg-boost-users_at_[hidden])
Date: 2002-04-10 16:34:41

<dick.bridges_at_[hidden]> wrote in message
news:OFAB9F41E0.6C6A5A14-ON88256B97.005045D8_at_tais.net...
>
> I'm implementing the Leader/Followers thread pool pattern - reference
> "Pattern-Oriented Software Architecture Volume 2" (Schmidt, et al).
>
> When a thread is released via notify_one(), it is [assumed to be] in the
> "leader" role and stores it's id.  A sanity check should be performed to
> verify that there is no other thread in the leader role (i.e.,
> current_leader is "none") before the thread stores it's id there.
>
> When one of the event sources is signaled and the leader is unblocked, it
> must, among other tasks, demote itself (i.e., set the current_leader value
> to "none")
> and promote one of the followers by notify_one() before continuing to
> service the request.  Because of the potential for race conditions,
> (especially in the context of future program updates, modifications, etc.)
> I should enforce the rule that only the leader can promote a follower.
> Something along the lines of:  current_leader == Thread::this().
>

I disagree (about the race condition).  Your condition variable is usually
used in conjunction with a mutex which protects your shared state's
invariants.

boost::mutex m;
boost::condition c;
Invariant:  leader != NONE

Pseudo-code for leader releasing

boost::mutex::scoped_lock l(m);    // lock mutex
if(TimeForANewLeader)
{
    leader=NONE;
    c.notify_one();        // Will signal one waiter.....

    // finish task - MUTEX IS STILL HELD!
    //
}

l.unlock();    // Only now will a waiter return from wait();

Pseudeo-code for follower waiting for signal:

boost::mutex::scoped_lock l(m);    // lock mutex

while(Not TimeForANewLeader)
{
    c.wait(m);    // Waiting automically releases AND reacquires
                        // the mutex
 }
// When we get here, we know we hold the mutex.
// Follower is responsible for "self-promotion"
leader=ME;

// Start my work as a leader.

An important note is that if we reach the while() statement and it IS time
for a new leader, then
we never wait, we already hold the mutex, and it is safe for us to promote
ourselves anyways!

Note that for the time being, you'll have to go outside of Boost.Threads for
the leader=ME statement because thread objects aren't copyable so
leader=boost::thread(); doesn't work to get the current thread.  Bill Kempf
has indicated that some form of portable thread_id s are coming in a future
version.

Hope this helps...
Dave