06_recap.en.md

Recap

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1. What was the most surprising for you? 😲

2. What was the most obvious for you? 🥱

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Points to remember

• Always use wait() with a predicate!
• Remember about spurious wake-ups and lost notifications
• Remember about a fight for locking a mutex when you use notify_all()
• You can't choose which thread should wake up when you use notify_one()

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Pre-test 🤯

// assume all necessary includes are here

int main() {
    std::mutex m;
    std::condition_variable cv;
    std::vector<int> v;
    std::vector<std::thread> producers;
    std::vector<std::thread> consumers;

    auto consume = [&] {
        std::unique_lock<std::mutex> ul(m);
        cv.wait(ul);
        std::cout << v.back();
        v.pop_back();
    };
    for (int i = 0; i < 10; i++) consumers.emplace_back(consume);

    auto produce = [&](int i) {
        {
            std::lock_guard<std::mutex> lg(m);
            v.push_back(i);
        }
        cv.notify_all();
    };
    for (int i = 0; i < 10; i++) producers.emplace_back(produce, i);

    for (auto && p : producers) p.join();
    for (auto && c : consumers) c.join();
}

1. there may be an Undefined Behavior in this code
2. the output is guaranteed to always be 0123456789
3. v is always an empty vector at the end of this program
4. if some producers threads started before some consumers, we could have a deadlock because of lost notifications
5. this code can be improved by providing a predicate to wait() to disallow getting elements when the vector is empty
6. a change from notify_all() to notify_one() + wait() with predicate will guarantee that each consumer thread will receive a different number

Note: 1, 4, 5, 6

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Post-work

• Ping-pong
  • difficult version - homework/ping_pong.cpp
  • easier version - homework/ping_pong_easier.cpp

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Homework: ping-pong

• Thread A prints "ping" and the consecutive number
• Thread B prints "pong" and the consecutive number
• Ping always starts. Pong always ends.
• Threads must work in turns. There may not be 2 consecutive pings or pongs. The program cannot end with a ping without a respective pong.
• The program must be terminated either after the given number of repetitions or after the time limit, whichever occurs first. The reason for termination should be displayed on the screen.
• Program parameters:
  • number of repetitions
  • time limit (in seconds)

$> g++ 03_ping_pong.cpp -lpthread
-std=c++17 -fsanitize=thread
$> ./a.out 1 10
ping 0
pong 0
Ping reached repetitions limit
Pong reached repetitions limit
$> ./a.out 12 1
ping 0
pong 0
ping 1
pong 1
ping 2
pong 2
Timeout

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Tips

If you got stuck:

• You need a mutex and a condition variable in your PingPong class
• Wait for a condition variable with wait_for() in stop() function
• Check the number of repetitions in ping and pong threads
• Use an additional std::atomic variable which will tell all threads to end, when the required conditions are met
• Ping and pong threads should be using wait() to check if it's their turn to work. Use an additional variable that will be used in the predicate passed to wait()
• The pong thread should end the program after reaching the repetition limit

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Repo on GitHub

https://github.com/coders-school/condition-variable

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