Volatile and multithreading: is the following thread-safe?

Question

Assume there are two threads running Thread1() and Thread2() respectively. The thread 1 just sets a global flag to tell thread 2 to quit and thread 2 periodically checks if it should quit.

volatile bool is_terminate = false;

void Thread1()
{
    is_terminate = true;
}

void Thread2()
{
    while (!is_terminate) {
        // ...
    }
}

I want to ask if the above code is safe assuming that access to is_terminate is atomic. I already know many materials state that volatile can not insure thread-safety generally. But in the situation that only one atomic variable is shared, do we really need to protect the shared variable using a lock?

Answer 1

It is probably sort of thread-safe.

Thread safety tends to depend on context. Updating a bool is always thread safe, if you never read from it. And if you do read from it, then the answer depends on when you read from it, and what that read signifies.

On some CPUs, but not all, writes to an object of type bool will be atomic. x86 CPUs will generally make it atomic, but others might not. If the update isn't atomic, then adding volatile won't help you.

But the next problem is reordering. The compiler (and CPU) will carry out reads/writes to volatile variables in the order specified, without any reordering. So that's good.

But it makes no guarantee about reordering one volatile memory access relative to all the non-volatile ones. So a common example is that you define some kind of flag to protect access to a resource, you make the flag volatile, and then the compiler moves the resource access up so it happens before you check the flag. It's allowed to do that, because it's not reordering the internal ordering of two volatile accesses, but merely a volatile and a non-volatile one.

Honestly, the question I'd ask is why not just do it properly? It is possible that volatile will work in this situation, but why not save yourself the trouble, and make it clearer that it's correct? Slap a memory barrier around it instead.