See the following classical producer-consumer code:
int main()
{
std::queue<int> produced_nums;
std::mutex m;
std::condition_variable cond_var;
bool done = false;
bool notified = false;
std::thread producer([&]() {
for (int i = 0; i < 5; ++i) {
std::this_thread::sleep_for(std::chrono::seconds(1));
std::unique_lock<std::mutex> lock(m);
std::cout << "producing " << i << '\n';
produced_nums.push(i);
notified = true;
cond_var.notify_one();
}
done = true;
cond_var.notify_one();
});
std::thread consumer([&]() {
std::unique_lock<std::mutex> lock(m);
while (!done) {
while (!notified) { // loop to avoid spurious wakeups
cond_var.wait(lock);
}
while (!produced_nums.empty()) {
std::cout << "consuming " << produced_nums.front() << '\n';
produced_nums.pop();
}
notified = false;
}
});
producer.join();
consumer.join();
}
I have copied this from cppreference.
Everything is pretty much straightforward to me, except the line in the consumer:
cond_var.wait(lock);
I do understand the loop that waits cond_var
to be notified, but why is it waiting for the lock?
cond_var.wait(lock);
does not wait for the lock.
That line does 3 things
lock
variablelock
variable again before it returns,It does all this atomically. While the thread is waiting for the condition variable, the mutex is not locked - that way your producer thread can acquire the lock and safely set any variables shared between the consumers/producer.
It locks the mutex again upon return, so the consumer again can safely access the shared variables.
If you tried to manage locking/unlocking the mutex yourself, you would end up with race conditions betwen locking/unlocking the mutex and waiting/signalling the condition variable - this is why waiting for a condition variable is tied to a mutex - so it can be done atomically, without race conditions.