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Enforcing Correct Mutex Usage with Synchronized Values


Anthony Williams is author of the bookC++ Concurrency in Action and of the just::thread C++0x thread library. He can be contacted at anthony.ajw@gmail.com.


In Associate Mutexes with Data to Prevent Races, Herb Sutter examined the importance of ensuring that the correct mutex is locked when accessing data, then presented one technique for achieving this. In this article, I present an alternative technique that I have used to achieve the same aims, in the form of the SynchronizedValue class template.

The Problem with Mutexes

The key problem with protecting shared data with a mutex is that there is no easy way to associate the mutex with the data. It is thus relatively easy to accidentally write code that fails to lock the right mutex --- or even locks the wrong mutex --- and the compiler will not help you.


std::mutex m1;
int value1;
std::mutex m2;
int value2;

int readValue1()
{
    std::lock_guard<std::mutex> lk(m1);
    return value1;
}
int readValue2()
{
    std::lock_guard<std::mutex> lk(m1); // oops: wrong mutex
    return value2;
}

Moreover, managing the mutex lock also clutters the source code, making it harder to see what is really going on.

The use of SynchronizedValue<T> solves both these problems --- the mutex is intimately tied to the value, so you cannot access it without a lock, and yet access semantics are still straightforward. For simple accesses, SynchronizedValue<T> behaves like a pointer-to-T; for example:


SynchronizedValue<std::string> value3;
std::string readValue3()
{
    return *value3;
}
void setValue3(std::string const& newVal)
{
    *value3=newVal;
}
void appendToValue3(std::string const& extra)
{
    value3->append(extra);
}

Both forms of pointer dereference return a proxy object rather than a real reference, to ensure that the lock on the mutex is held across the assignment or method call, but this is transparent to the user.

Beyond Simple Accesses

The pointer-like semantics work very well for simple accesses such as assignment and calls to member functions. However, sometimes you need to perform an operation that requires multiple accesses under protection of the same lock, and that's what the update() method provides.

By calling update() you obtain an Updater object that holds a lock on the mutex protecting the data, and which can be used to access the protected data. The lock is held until the Updater object is destroyed, so you can safely perform multi-part operations. The Updater object also acts as a pointer-to-T, just like SynchronizedValue does, but this time the lock is already held. For example, the following function adds a trailing slash to a path held in a SynchronizedValue<std::string>. The use of the Updater object ensures that the string hasn't changed in between the query and the update.


void addTrailingSlashIfMissing(SynchronizedValue<std::string> & path)
{
    SynchronizedValue<std::string>::Updater u=path.update();

    if(u->empty() || (*u->rbegin()!='/'))
    {
        *u+='/';
    }
}

Operations Across Multiple Objects

Though SynchronizedValue<T> works very well for protecting a single object of type T, nothing that we've seen so far solves the problem of operations that require atomic access to multiple objects unless those objects can be combined within a single structure protected by a single mutex.

One way to protect access to two SynchronizedValue<T> objects is to construct a SynchronizedValue<T>::Updater for each object and use those to access the respective protected values; for instance:


SynchronizedValue<std::queue<MessageType> > q1,q2;
void transferMessage()
{
    SynchronizedValue<std::queue<MessageType> >::Updater u1=q1.update();
    SynchronizedValue<std::queue<MessageType> >::Updater u2=q2.update();

    if(!u1->empty())
    {
        u2->push_back(u1->front());
        u1->pop_front();
    }
}

This works well in some scenarios, but not all -- if the same two objects are updated together in different sections of code then you need to take care to ensure that the Updater objects are constructed in the same sequence in all cases, otherwise you have the potential for deadlock. This is just the same as when acquiring any two mutexes.


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