Question

I want to use a shared mutex so threads only get locked when a vector/map/whatever is written to rather than read from. But I think func2() will never get the uniqueue lock because func1() will never get to unlock. Is there any way to not count a same-thread lock on shared_mutex when trying to get the uniqueue lock? Or would the problem still occur even then?

I’m guessing I need to find a way to force-get the lock (one thread at a time) once all threads have reached func2() OR have released the lock.

func2()
{
    boost::unique_lock<boost::shared_mutex> lock_access3(shared_mutex);
    /*stuff*/
    lock_access3.unlock(); 
}

func1()
{
    boost::shared_lock<boost::shared_mutex> lock_access1(shared_mutex);
    func2();
    lock_access1.unlock();
}

Answer 1

Two things you need to do:

1) Since func1 implements a write operation, it needs to acquire an exclusive lock, not a shared lock.

2) Since func2 is called with the lock already held, it should not try to acquire the lock.