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I created a SafeQueue class, which stores pointers. It has two methods:

  • push: Adds a new pointer to the queue
  • next: If the queue is empty, returns nullptr. Otherwise it returns the front element, and pop the queue

Most of the time, I have one producer and one consumer. But there may be more.

The producer just simply calls .push(ptr)

The consumer(s) call .next(), until a nullptr is returned. Or they continue the loop forever.

The reason I did this, is to eliminate .isEmpty function, so between the .empty() and .front() no other thread can pop the queue.

It works, but I don't think, that it is the preferable or an optimal solution for the problem.

The code is:

template<class T>
class SafeQueue {

    std::queue<T*> q;
    std::mutex m;

public:

    SafeQueue() {}

    void push(T* elem) {

        m.lock();
        if(elem != nullptr) {
            q.push(elem);
        }
        m.unlock();

    }

    T* next() {

        T* elem = nullptr;

        m.lock();
        if(!q.empty()) {
            elem = q.front();
            q.pop();
        }
        m.unlock();

        return elem;

    }

};
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2 Answers 2

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I see some things that may help you improve your code.

List all required #includes

The code needs the following #includes to actually compile. Since they therefore form part of the interface, they should be included in the file and in a code review:

#include <mutex>
#include <queue>

Be clear about ownership

If the intention is to have a thread-safe queue, then passing pointers in and out is definitely not the way to go. The problem is with object ownership. Even if your thread-safe queue works perfectly, all of its inherent goodness is all too easily bypassed because you're using pointers. For example:

SafeQueue<std::string> sq;

{ 
    std::string msg1{"this message exists"};
    sq.push(&msg1);
}   // msg1 is now destroyed, but queue still has pointer
std::cout << *sq.next() << " no longer\n";  // kaboom!

The problem is that the queue doesn't actually own the object (or at least have a std::shared_ptr) so there's not much point in perfecting the queue until that's addressed.

Choose better names

I have never thought of push and next as inverse operations, and I'll bet you never have either. The push member function name is OK, since it actually does that, but next is just a strange name. I'd say use pop or pop_front might be better names.

Minimize locking duration

In the push code, we have this:

void push(T* elem) {
    m.lock();
    if(elem != nullptr) {
        q.push(elem);
    }
    m.unlock();
}

But why acquire a lock if you don't need it? It just slows things down. You could instead write that like this:

void push(T* elem) {
    if (elem == nullptr) {
        return;
    }
    m.lock();
    q.push(elem);
    m.unlock();
}

Or better, see the next suggestion:

Use RAII to reduce errors

If you happened to forget to remove the lock on the code above, Bad Things would likely happen. Fortunately, in C++, there's a handy idiom that is very often used in C++ and it's called Resource Allocation is Initialization. In this context, we use a std::lock_guard like this:

void push(T elem) {
    std::lock_guard<std::mutex> lock(m);
    q.push(elem);
}

The lock_guard automatically gets locked on creation and unlocked on deletion, so when it goes out of scope, the lock is released even if you forget.

Return an indicator of success

Since we've already established that passing pointers in or out is a problem, I'd suggest changing the interface for the next() function. Have it take a reference (so the caller must supply one) and then return a bool to indicate success. That might look like this:

bool next(T& elem) {
    std::lock_guard<std::mutex> lock(m);
    if (q.empty()) {
        return false;
    }
    elem = q.front();
    q.pop();
    return true;
}
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1
  • 1
    \$\begingroup\$ Good point about the minimize lock duration. \$\endgroup\$
    – Surt
    Dec 13, 2016 at 17:10
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For this kind of queue I often use something like this

std::mutex mtx;
std::condition_variable cv;

void push(T* elem) {
    if(elem == nullptr) {
        return;
    }
    std::unique_lock<std::mutex> lck(mtx);
    q.push(elem);
    cv.notify_one();
}

T* next() {
    T* elem = nullptr;

    std::unique_lock<std::mutex> lck(mtx);
    cv.wait(lck, !q.empty());

    if(!q.empty()) {
        elem = q.front();
        q.pop();
    }
   
    return elem;
}

The consumers lock on the empty queue, and when something is pushed one is awakened for each push'ed. That way you don't waste CPU cycles waiting when you don't have anything to process, but there is a certain cost to locking and the task switch.

So measure your performance.

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  • \$\begingroup\$ Why did you use unique_lock in the push method too? Isn't it only required in the next function? Is it possible to use a lock_guard instead of unique_lock in the push ? \$\endgroup\$
    – Iter Ator
    Dec 13, 2016 at 10:49
  • \$\begingroup\$ @IterAtor easy answer, habit, using the same lock style make me feel safe. There are some requirement on the CV lock, which I couldn't remember required using the same lock in other places. \$\endgroup\$
    – Surt
    Dec 13, 2016 at 17:08
  • 1
    \$\begingroup\$ You always have to pass a lambda or a function to std::condition_variable::wait so the expression gets evaluated when a spuriously wakeup happens. \$\endgroup\$
    – JulianW
    Mar 30, 2019 at 9:16
  • \$\begingroup\$ in your push() you need to use a shared_lock, not a unique_lock. The unique_lock is a deferred lock. \$\endgroup\$
    – edwinc
    Aug 3, 2020 at 19:03
  • \$\begingroup\$ @edwinc I can't find a place where it says std::defer_lock is the default, could you add a reference please. \$\endgroup\$
    – Surt
    Aug 7, 2020 at 11:54

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