std::shared_mutex

#include <iostream>
#include <mutex>  // For std::unique_lock
#include <shared_mutex>
#include <thread>
 
class ThreadSafeCounter {
 public:
  ThreadSafeCounter() = default;
 
  // Multiple threads/readers can read the counter's value at the same time.
  unsigned int get() const {
    std::shared_lock lock(mutex_);
    return value_;
  }
 
  // Only one thread/writer can increment/write the counter's value.
  void increment() {
    std::unique_lock lock(mutex_);
    value_++;
  }
 
  // Only one thread/writer can reset/write the counter's value.
  void reset() {
    std::unique_lock lock(mutex_);
    value_ = 0;
  }
 
 private:
  mutable std::shared_mutex mutex_;
  unsigned int value_ = 0;
};
 
int main() {
  ThreadSafeCounter counter;
 
  auto increment_and_print = [&counter]() {
    for (int i = 0; i < 3; i++) {
      counter.increment();
      std::cout << std::this_thread::get_id() << ' ' << counter.get() << '\n';
 
      // Note: Writing to std::cout actually needs to be synchronized as well
      // by another std::mutex. This has been omitted to keep the example small.
    }
  };
 
  std::thread thread1(increment_and_print);
  std::thread thread2(increment_and_print);
 
  thread1.join();
  thread2.join();
}
 
// Explanation: The output below was generated on a single-core machine. When
// thread1 starts, it enters the loop for the first time and calls increment()
// followed by get(). However, before it can print the returned value to
// std::cout, the scheduler puts thread1 to sleep and wakes up thread2, which
// obviously has time enough to run all three loop iterations at once. Back to
// thread1, still in the first loop iteration, it finally prints its local copy
// of the counter's value, which is 1, to std::cout and then runs the remaining
// two loop iterations. On a multi-core machine, none of the threads is put to
// sleep and the output is more likely to be in ascending order.