How to assert if a std::mutex is locked?

Question

With GCC 4.8.2 (on Linux/Debian/Sid 64 bits) -or GCC 4.9 when available - in C++11- I have some mutex

std::mutex gmtx;

^{actually, it is}

Answer 1

Strictly speaking, the question was about checking the lockedness of std::mutex directly. However, if encapsulating it in a new class is allowed, it's very easy to do so:

class mutex :
    public std::mutex
{
public:
#ifndef NDEBUG
    void lock()
    {
        std::mutex::lock();
        m_holder = std::this_thread::get_id(); 
    }
#endif // #ifndef NDEBUG

#ifndef NDEBUG
    void unlock()
    {
        m_holder = std::thread::id();
        std::mutex::unlock();
    }
#endif // #ifndef NDEBUG

#ifndef NDEBUG
    /**
    * @return true iff the mutex is locked by the caller of this method. */
    bool locked_by_caller() const
    {
        return m_holder == std::this_thread::get_id();
    }
#endif // #ifndef NDEBUG

private:
#ifndef NDEBUG
    std::atomic<std::thread::id> m_holder;
#endif // #ifndef NDEBUG
};

Note the following:

1. In release mode, this has zero overhead over std::mutex except possibly for construction/destruction (which is a non-issue for mutex objects).
2. The m_holder member is only accessed between taking the mutex and releasing it. Thus the mutex itself serves as the mutex of m_holder. With very weak assumptions on the type std::thread::id, locked_by_caller will work correctly.
3. Other STL components, e.g., std::lock_guard are templates, so they work well with this new class.

Answer 2

You could just use a recursive_mutex, which can be locked multiple times on the same thread. Note: If it were my code, I would restructure it so that I don't need a recursive_mutex, but it will address your problem.

Answer 3

It's not technically an assertion, but I've used a similar approach to prevent unlocked access to shared state: add a reference parameter to the lock guard class on the in the unsafe function (beta in your example). Then the function cannot be called unless the caller has created a lock guard. It solves the problem of accidentally calling the function outside of the lock, and it does so at compile time with no races.

So, using your example:

typedef std::lock_guard<std::mutex> LockGuard;
void alpha(void) {
   LockGuard g(gmtx);
   beta(g);
   // some other work
}

void beta(LockGuard&) {
   // some real work
}

void gamma(void) {
   LockGuard g(gmtx);
   beta(g);
   // some other work
}

//works recursively too
void delta(LockGuard& g)
{
   beta(g);
}

Drawbacks:

• Does not validate that the lock actually wraps the correct mutex.
• Requires the dummy parameter. In practice, I usually keep such unsafe functions private so this is not an issue.

Answer 4

My solution is simple, use try_lock to test then unlock if needed:

std::mutex mtx;

bool is_locked() {
   if (mtx.try_lock()) {
      mtx.unlock();
      return false;
   }
   return true; // locked thus try_lock failed
}

Answer 5

Try atomic (e.g. atomic<bool> or atomic<int>), which has a nice load function that will do what you want, as well as other nice functions like compare_exchange_strong.

Answer 6

std::unique_lock<L> has owns_lock member function (equivalent of is_locked as you say).

std::mutex gmtx;
std::unique_lock<std::mutex> glock(gmtx, std::defer_lock);

void alpha(void) {
   std::lock_guard<decltype(glock)> g(glock);
   beta(void);
   // some other work
}
void beta(void) {
   assert(glock.owns_lock()); // or just assert(glock);
   // some real work
}

EDIT: In this solution, all lock operations should be performed via unique_lock glock not 'raw' mutex gmtx. For example, alpha member function is rewritten with lock_guard<unique_lock<mutex>> (or simply lock_guard<decltype(glock)>).