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Q&A Destroy std::mutex referenced but not owned by std::unique_lock?

Is it correct to destroy a mutex which is referenced but not owned by an unique_lock as in this code? { std::unique_ptr<std::mutex> mutex = std::make_unique<std::mutex>(); std::u...

1 answer  ·  posted 3y ago by Estela‭  ·  last activity 3y ago by Angew‭

Question c++
#2: Post edited by user avatar Estela‭ · 2021-02-04T09:35:39Z (about 3 years ago)
  • Is it correct to destroy a mutex which is referenced but not owned by an unique_lock as in this code?
  • {
  • std::unique_ptr<std::mutex> mutex = std::make_unique<std::mutex>();
  • std::unique_lock<std::mutex> lock(*mutex);
  • lock.unlock();
  • mutex.reset();
  • } // Here lock is destroyed while holding an invalid reference to *mutex
  • // but not having a lock on any mutex at all.
  • The example is somewhat convoluted and for this simple case it would be easier to not `lock.unlock();mutex.reset()` and let the destructors do their work. But there are other situations where being able to do this might come handy.
  • It can be read in [std::unique_lock<Mutex>::~unique_lock](https://en.cppreference.com/w/cpp/thread/unique_lock/%7Eunique_lock) that :
  • `Destroys the lock. If *this has an associated mutex and has acquired ownership of it, the mutex is unlocked.
  • `
  • Which leads me to believe that if the unique_lock no longer has ownership of the mutex the mutex is not unlocked so the no longer valid reference the mutex is not used in the destructor.
  • Is it correct to destroy a mutex which is referenced but not owned by an unique_lock as in this code?
  • ```c++
  • {
  • std::unique_ptr<std::mutex> mutex = std::make_unique<std::mutex>();
  • std::unique_lock<std::mutex> lock(*mutex);
  • lock.unlock();
  • mutex.reset();
  • } // Here lock is destroyed while holding an invalid reference to *mutex
  • // but not having a lock on any mutex at all.
  • ```
  • The example is somewhat convoluted and for this simple case it would be easier to not `lock.unlock();mutex.reset()` and let the destructors do their work. But there are other situations where being able to do this might come handy.
  • It can be read in [std::unique_lock<Mutex>::~unique_lock](https://en.cppreference.com/w/cpp/thread/unique_lock/%7Eunique_lock) that :
  • `Destroys the lock. If *this has an associated mutex and has acquired ownership of it, the mutex is unlocked.
  • `
  • Which leads me to believe that if the unique_lock no longer has ownership of the mutex the mutex is not unlocked so the no longer valid reference the mutex is not used in the destructor.
#1: Initial revision by user avatar Estela‭ · 2021-02-04T09:33:37Z (about 3 years ago)
Destroy std::mutex referenced but not owned by std::unique_lock?
Is it correct to destroy a mutex which is referenced but not owned by an unique_lock as in this code?

    {
      std::unique_ptr<std::mutex> mutex = std::make_unique<std::mutex>();
      std::unique_lock<std::mutex> lock(*mutex);
      lock.unlock();
      mutex.reset();
    } // Here lock is destroyed while holding an invalid reference to *mutex
      // but not having a lock on any mutex at all.
The example is somewhat convoluted and for this simple case it would be easier to not `lock.unlock();mutex.reset()` and let the destructors do their work. But there are other situations where being able to do this might come handy.

It can be read in [std::unique_lock<Mutex>::~unique_lock](https://en.cppreference.com/w/cpp/thread/unique_lock/%7Eunique_lock) that :

`Destroys the lock. If *this has an associated mutex and has acquired ownership of it, the mutex is unlocked.
`

Which leads me to believe that if the unique_lock no longer has ownership of the mutex the mutex is not unlocked so the no longer valid reference the mutex is not used in the destructor.
c++