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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...
Question
c++
#2: Post edited
by
Estela
·
2021-02-04T09:35:39Z (almost 4 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Which leads me to believe that if the unique_lock no longer has ownership of the mutex the mutex is not unlocked so the no longer valid reference the mutex is not used in the destructor.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Which leads me to believe that if the unique_lock no longer has ownership of the mutex the mutex is not unlocked so the no longer valid reference the mutex is not used in the destructor.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
Estela
·
2021-02-04T09:33:37Z (almost 4 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.