How to use a std::mutex in a class context

自古美人都是妖i 提交于 2019-12-03 13:33:06

std::mutex is neither copyable nor movable. Including one in your class necessitates that your class becomes non-copyable (-movable) as well. If you want your class to be copyable or movable, you will have to tell the compiler how objects of your class are to be copied or moved by implementing copy/move construction/assignment yourself. For example:

class synchronized_int {
  mutable std::mutex mtx;
  int value;
public:
  synchronized_int(int v = 0) : value(v) {}

  // Move initialization
  synchronized_int(synchronized_int&& other) {
    std::lock_guard<std::mutex> lock(other.mtx);
    value = std::move(other.value);
    other.value = 0;
  }

  // Copy initialization
  synchronized_int(const synchronized_int& other) {
    std::lock_guard<std::mutex> lock(other.mtx);
    value = other.value;
  }

  // Move assignment
  synchronized_int& operator = (synchronized_int&& other) {
    std::lock(mtx, other.mtx);
    std::lock_guard<std::mutex> self_lock(mtx, std::adopt_lock);
    std::lock_guard<std::mutex> other_lock(other.mtx, std::adopt_lock);
    value = std::move(other.value);
    other.value = 0;
    return *this;
  }

  // Copy assignment
  synchronized_int& operator = (const synchronized_int& other) {
    std::lock(mtx, other.mtx);
    std::lock_guard<std::mutex> self_lock(mtx, std::adopt_lock);
    std::lock_guard<std::mutex> other_lock(other.mtx, std::adopt_lock);
    value = other.value;
    return *this;
  }
};

Declare your mutex dynamically and not statically.

At your Database class header declare the mutex as a pointer, like that:

mutex * semaphore;

and at your constructor initialize the mutex calling his constructor:

semaphore = new std::mutex();
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