I've been using boost::mutex::scoped_lock
in this manner:
void ClassName::FunctionName()
{
{
boost::mutex::scoped_lock scopedLock(mutex_);
//do stuff
waitBoolean=true;
}
while(waitBoolean == true ){
sleep(1);
}
//get on with the thread's activities
}
Basically it sets waitBoolean, and the other thread signals that it is done by setting waitBoolean to false;
This doesn't seem to work, however, because the other thread can't get a lock on mutex_ !!
I was assuming that by wrapping the scoped_lock in brackets I would be terminating its lock. This isn't the case? Reading online says that it only gives up the mutex when the destructor is called. Won't it be destroyed when it goes out of that local scope?
Signaling part of code:
while(running_){
boost::mutex::scoped_lock scopedLock(mutex_);
//Run some function that need to be done...
if(waitBoolean){
waitBoolean=false;
}
}
Thanks!
To synchronize two threads use a condition variable. That is the state of the art way to synchronize two threads the way you want :
Using boost, the waiting part is something like :
void BoostSynchronisationPoint::waitSynchronisation()
{
boost::unique_lock<boost::mutex> lock(_mutex);
_synchronisationSent = false;
while(!_synchronisationSent)
{
_condition.wait(lock); // unlock and wait
}
}
The notify part is something like :
void BoostSynchronisationPoint::sendSynchronisation()
{
{
boost::lock_guard<boost::mutex> lock(_mutex);
_synchronisationSent = true;
}
_condition.notify_all();
}
The business with _synchronisationSent is to avoid spurrious wakeups : see wikipedia
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