std::atomic to what extent?

Question

In the C++ Seasoning video by Sean Parent https://youtu.be/W2tWOdzgXHA at 33:41 when starting to talk about “no raw synchronization primitives”, he brings an example to show that with raw synchronization primitives we will get it wrong. The example is a bad copy on write class:

template <typename T>
class bad_cow {
    struct object_t {
        explicit object_t(const T& x) : data_m(x) { ++count_m; }
        atomic<int> count_m;
        T data_m;
    };
    object_t* object_m;

public:
    explicit bad_cow(const T& x) : object_m(new object_t(x)) { }
    ~bad_cow() { if (0 == --object_m->count_m) delete object_m; }
    bad_cow(const bad_cow& x) : object_m(x.object_m) { ++object_m->count_m; }

    bad_cow& operator=(const T& x) {
        if (object_m->count_m == 1) {
            // label #2
            object_m->data_m = x; 
        } else {
            object_t* tmp = new object_t(x);
            --object_m->count_m; // bug #1
            // this solves bug #1:
            // if (0 == --object_m->count_m) delete object_m;
            object_m = tmp;
        }
        return *this;
    }
};

He then asks the audience to find the bug, which is the bug #1 as he confirms.

But a more obvious bug I guess, is when some thread is about to proceed to execute a line of code that I have denoted with label #2, while all of a sudden, some other thread just destroys the object and the destructor is called, which deletes object_m. So, the first thread will encounter a deleted memory location.

Am I right? I don’t seem so!

Is it just me or does count_m have an undefined value when object_t is constructed. The constructor for std::atomic<int> does not initialize the underlying value to zero - at least not on VS2017 or VS2019. — selbie
– selbie, Commented Nov 28, 2019 at 23:04

Jeremy Friesner · Accepted Answer · 2019-11-28 22:35:37Z

some other thread just destroys the object and the destructor is called, which deletes object_m. So, the first thread will encounter a deleted memory location.

Am I right? I don’t seem so!

Assuming the rest of the program isn't buggy, that shouldn't happen, because each thread should have its own reference-count object referencing the data_m object. Therefore, if thread B has a bad_cow object that references the data-object, then thread A cannot (or at least should not) ever delete that object, because the count_m field can never drop to zero as long as there remains at least one reference-count object pointing to it.

Of course, a buggy program might encounter the race condition you suggest -- for example, a thread might be holding only a raw pointer to the data-object, rather than a bad_cow that increments its reference count; or a buggy thread might call delete on the object explicitly rather than relying on the bad_cow class to handle deletion properly.

6502 · Accepted Answer · 2019-11-29 07:22:31Z

Your objection doesn't hold because *this at that moment is pointing to the object and the count is 1. The counter cannot get to 0 unless someone is not playing this game correctly (but in that case anything can happen anyway).

Another similar objection could be that while you're assigning to *this and the code being executed is inside the #2 branch another thread makes a copy of *this; even if this second thread is just reading the pointed object may see it mutating suddenly because of the assignment. The problem in this case is that count was 1 when entering the if in the thread doing the mutation but increased immediately after.

This is also however a bad objection because this code handles concurrency to the pointed-to object (like for example std::shared_ptr does) but you are not allowed to mutate and read a single instance of bad_cow class from different threads. In other words a single instance of bad_cow cannot be used from multiple threads if some of them are writers without adding synchronization. Distinct instances of bad_cow pointing to the same storage are instead safe to be used from different threads (after the fix #1, of course).

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