Question

Is unique_ptr thread safe? Is it impossible for the code below to print same number twice?

#include <memory>
#include <string>
#include <thread>
#include <cstdio>

using namespace std;

int main()
{
    unique_ptr<int> work;

    thread t1([&] {
        while (true) {
            const unique_ptr<int> localWork = move(work);
            if (localWork)
                printf("thread1: %d\n", *localWork);
            this_thread::yield();
        }
    });

    thread t2([&] {
        while (true) {
            const unique_ptr<int> localWork = move(work);
            if (localWork)
                printf("thread2: %d\n", *localWork);
            this_thread::yield();
        }
    });

    for (int i = 0; ; i++) {
        work.reset(new int(i));

        while (work)
            this_thread::yield();
    }

    return 0;
}

Answer 1

No, it isn’t thread-safe.

Both threads can potentially move the work pointer with no explicit synchronization, so it’s possible for both threads to get the same value, or both to get some invalid pointer … it’s undefined behaviour.

If you want to do something like this correctly, you probably need to use something like std::atomic_exchange so both threads can read/modify the shared work pointer with the right semantics.