c++ atomic read/write misunderstanding

Question

Why program with this code sometimes prints "2" ?

int main() {
    std::atomic<int> a;
    a = 0;

    std::thread t1([&]{++a;});
    std::thread t2([&]{a++;});
    std::thread t3([&]{
        a = a.load() + 1;
    });

    t1.join();
    t2.join();
    t3.join();

    if (a != 3) {
        std::cout << "a: " << a << std::endl;
    }
}

I've thought std::atomic guarantees that all operations will be done atomically so writing here(incrementing) will use a memory barrier and we will have always 3 at the end of threads work. I've explored the code and found out that the problem thread is t3 but I can't understand why it is wrong.

Answer 1

t3, unlike the two other threads, does not perform an atomic add. Instead, it atomically loads a, performs the arithmetic (add 1) on a temporary, and atomically stores that new value back to a. This overwrites a regardless of atomic operations that might have happened in between.

So you can have the following scenario:

1. t1 or t2 atomically increment a which is now equal to 1.
2. t3 atomically loads 1.
3. t1 or t2 atomically increment a which is now equal to 2.
4. t3 performs a non-atomic add on the previously loaded value and atomically stores back 2.