i\'m trying to create a new thread with a class \"CameraManager\" but i have the following error:
cannot convert \'*void(CameraManager:: * )(void*) to
If you want the function to be a member of the class, it must be static
. It's because the thread function will be called directly and will not have a valid this
pointer. This can be solved by having a wrapper function, that gets passed the actual object and then calls the proper member function:
void *dequeueLoopWrapper(void *p)
{
CameraManager *cameraManager = static_cast<CameraManager*>(p);
camereraManager->dequeueLoop();
return nullptr;
}
// ...
void CameraManager::startDequeuing()
{
dequeuing = true;
dequeueThreadId = pthread_create(&dequeueThread, NULL, dequeueLoopWrapper, this);
}
However, I would recommend you start using the threading support in the new standard library:
void CameraManager::startDequeuing()
{
dequeuing = true;
myThread = std::thread(&CameraManager::dequeueLoop, this);
}
I don't want to declare
dequeueLoop
as a static function
If you want to use pthreads, then you'll need a static or non-member function for the entry point. You can pass a pointer to your object to this function, using it as a trampoline into the non-static member function:
static void * dequeueEntry(void * self) {
return static_cast<CameraManager*>(self)->dequeueLoop();
}
dequeueThreadId = pthread_create(
&dequeueThread, NULL,
&CameraManager::dequeueEntry, // <-- pointer to trampoline function
this); // <-- pointer to object for member function
Alternatively, if you have a modern compiler, you could use the standard thread library instead:
std::thread thread(&CameraManager::dequeLoop, this);
You can't use a pointer to member function as a function pointer unless it's static. You'll have to make dequeueLoop a free function, or write a free function as a wrapper to it.
To access the class members in a free function, you should have the function pass it's this
pointer as the final argument of pthread_create. Then have the free function cast it's argument to a pointer to the class.