CODEWITHSUNDEEP
c++queuethread-safetymutex
Eagle06
Eagle06

Reputation: 71

Is the following std::queue code example thread safe?

I have the following code example and I'm not sure if it is thread safe.

void remove(/*...*/) {
    int a;
    float* data;

    while (true)
    {
        unique_lock<mutex> lock1(mu);
        cond.wait(lock1, [this](){return count > 0;});
        lock1.unlock();

        a = std_queue1.front();

        data = std_queue2.front();

        // process data

        unique_lock<mutex> lock2(mu);
        std_queue1.pop();
        std_queue2.pop();
        count--;
        lock2.unlock();
        cond.notify_all();
    }
}

void add(/*...*/){
    int a, iterations = 0;
    vector<float*> data;
    // Allocate some data entries
    while (true) {
        // set a
        unique_lock<mutex> lock1(mu);
        std_queue1.push(a);
        cond.wait(lock1, [this, &limit]() { return count < limit; });
        lock1.unlock();

        // Init data

        if (iterations++ > 2) {
            unique_lock<mutex> lock2(mu);
            std_queue2.push(data[/*...*/]);
            count++;
            lock2.unlock();
            cond.notify_all();
        }
    }
}

remove(...) and add(...) are functions of the same class. Each function is executed by one thread and both Threads access the same queues. A mutex is locked to push and pop data of the queues, but not for the function front().

So my question is: Is this example thread safe? Or do I need a mutex for the front() function? And is the mutex necessary for the functions pop() and push()?

Upvotes: 0

Views: 146

Answers (1)

eerorika
eerorika

Reputation: 238311

So my question is: Is this example thread safe?

No.

Or do I need a mutex for the front() function?

Yes.

And is the mutex necessary for the functions pop() and push()?

Yes.

Modifying a standard container is not guaranteed to be thread safe.

Upvotes: 2

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