使用互斥量将一次运行的线程数限制为2

Question

I have a program that pushes 10 threads into a vector, each of which is supposed to print out a character 5 times before finishing ('A' for the first thread, 'B' for the second, etc). I'm able to get them to either run all at once (using detach()) or have them run one at a time (using join()). Now I want to use a Mutex to limit the number of threads allowed to print at a time to 2. I've been able to declare the mutex and put the lock in place but I'm unsure of how to apply a limit like this. Anyone have any ideas on how to proceed?

deque<int> q ;
mutex print_mutex ;
mutex queue_mutex ;
condition_variable queue_cond ;

void begin(int num) {
    unique_lock<mutex> ul {queue_mutex};
    q.emplace_back(num);
    queue_cond.wait(ul,[num]{
        return q.front() == num; });
    q.pop_front();
    cout << num << " leaves begin " << endl ;
}

void end ( int num ) {
    lock_guard<mutex>lg{queue_mutex};
    queue_cond.notify_all();
    cout << num << " has ended " << endl ;
}

void run(int num, char ch) {
    begin(num);
    for (int i = 0; i < 5; ++i) {
        {
            lock_guard<mutex> lg { print_mutex };
            cout << ch << endl << flush ;
        }
        sleep_for(milliseconds(250));
    }
    end(num);
}

int main() {
    vector<thread>threads {};
    for (int i = 0; i < 10; ++i) {
        threads.push_back(thread{run,i,static_cast<char>(65+i)});
        threads.at(i).join();
    }
}

Answer 1

You have already set up a FIFO for your threads with the global deque<int> q . So let's use that.

Currently, you're trying to restrict execution until the current thread is at the front. Although there's a bug, because begin will immediately pop that thread from the deque. Better to remove the value when you call end . Here's that change, first:

void end(int num)
{
    {
        lock_guard<mutex>lg{queue_mutex};
        cout << num << " has ended " << endl ;
        q.erase(find(q.begin(), q.end(), num));
    }
    queue_cond.notify_all();
}

This uses std::find from <algorithm> to remove the specific value. You could use pop_front , but we're about to change that logic so this is more generic. Also notice you don't need to lock the condition variable when you notify.

So, it's not much of a stretch to extend the logic in begin to be in the first two places. Here:

void begin(int num)
{
    unique_lock<mutex> ul {queue_mutex};
    q.emplace_back(num);
    queue_cond.wait(ul,[num]{
        auto end = q.begin() + std::min(2, static_cast<int>(q.size()));
        return find(q.begin(), end, num) != end;
        });
    cout << num << " leaves begin " << endl ;
}

You can change that 2 to anything you want, allowing up to that many threads to pass. At some point, you would probably abandon this approach and use something simpler like a single counter variable, then rely on the thread scheduler to manage which thread is woken, rather than force them into your FIFO. That way you can switch to using notify_one to wake a single thread and reduce switching overhead.

Anyway, the last thing to do is remove the join from your thread generation loop. The concurrency is now managed by begin and end . So you would do this:

for (int i = 0; i < 10; ++i) {
    threads.push_back( thread{run, i, 'A'+i} );
}
for (auto & t : threads) t.join();