Do locks in C++ 11 guarantee freshness of accessed data?

Question

Usually, when using std::atomic types accessed concurrently by multiple threads, there's no guarantee a thread will read the "up to date" value when accessing them, and a thread may get a stale value from cache or any older value. The only way to get the up to date value are functions such as compare_exchange_XXX. (See questions here and here)

#include <atomic>

std::atomic<int> cancel_work = 0;
std::mutex mutex;

//Thread 1 executes this function
void thread1_func() 
{
    cancel_work.store(1, <some memory order>);
}


// Thread 2 executes this function
void thread2_func() 
{
    //No guarantee tmp will be 1, even when thread1_func is executed first
   int tmp = cancel_work.load(<some memory order>); 
}

However my question is, what happens when using a mutex and lock instead? Do we have any guarantee of the freshness of shared data accessed?

For example, assuming both thread 1 and thread 2 are run concurrently and thread 1 obtains the lock first (executes first). Does it guarantee that thread 2 will see the modified value and not an old value? Does it matter whether the shared data "cancel_work" is atomic or not in this case?

#include <atomic>

int cancel_work = 0;  //any difference if replaced with std::atomic<int> in this case?
std::mutex mutex;

// Thread 1 executes this function
void thread1_func() 
{
    //Assuming Thread 1 enters lock FIRST
    std::lock_guard<std::mutex> lock(mutex);

    cancel_work = 1;
}


// Thread 2 executes this function
void thread2_func() 
{
    std::lock_guard<std::mutex> lock(mutex);

    int tmp = cancel_work; //Will tmp be 1 or 0?
}

int main()
{
   std::thread t1(thread1_func);
   std::thread t2(thread2_func);

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

   return 0;
}

Answer 1

Yes, the using of the mutex/lock guarantees that thread2_func() will obtain a modified value.

However, according to the std::atomic specification:

The synchronization is established only between the threads releasing and acquiring the same atomic variable. Other threads can see different order of memory accesses than either or both of the synchronized threads.

So your code will work correctly using acquire/release logic, too.

#include <atomic>

std::atomic<int> cancel_work = 0;

void thread1_func() 
{
    cancel_work.store(1, std::memory_order_release);
}

void thread2_func() 
{
    // tmp will be 1, when thread1_func is executed first
    int tmp = cancel_work.load(std::memory_order_acquire); 
}

Answer 2

The C++ standard only constrains the observable behavior of the abstract machine in well formed programs without undefined behavior anywhere during the abstract machine's execution.

It provides no guarantees about mapping between the physical hardware actions the program executes and behavior.

In your cases, on the abstract machine there is no ordering between thread1 and thread2's execution. Even if the physical hardware where to schedule and run thread1 before thread2, that places zero constraints (in your simple example) on the output the program generates. The programs' output is only contrained by what legal outputs the abstract machine could produce.

A C++ compiler can legally:

1. Eliminate your program completely as equivalent to return 0;

2. Prove that the read of cancel_work in thread2 is unsequenced relative to all modification of cancel_work away from 0, and change it to a constant read of 0.

3. Actually run thread1 first then run thread2, but prove it can treat the operations in thread2 as-if they occurred before thread1 ran, so don't bother forcing a cache line refresh in thread2 and reading stale data from cancel_work.

What actually happens on the hardware does not impact what the program can legally do. And what the program can legally do is in threading sitations is restricted by observable behavior of the abstract machine, and on the behavior of synchronization primitives and their use in different threads.

For an actual happens before relationship to occur, you need something like:

std::thread(thread1_func).join();
std::thread(thread2_func).join();

and now we do know that everything in thread1_func happens before thread2_func.

We can still rewrite your program as return 0; and similar changes. But we now have a guarantee that thread1_func happens before thread2_func code does.

Note that we can eliminate (1) above via:

std::lock_guard<std::mutex> lock(mutex);

int tmp = cancel_work; //Will tmp be 1 or 0?
std::cout << tmp;

and cause tmp to actually be printed.

The program can then be converted to one that prints 1 or 0 and has no threading at all. It could keep the threading, but change thread2_func to print a constant 0. Etc.

---

So we rewrite your program to look like this:

std::condition_variable cv;
bool writ = false;
int cancel_work = 0;  //any difference if replaced with std::atomic<int> in this case?
std::mutex mutex;

// Thread 1 executes this function
void thread1_func() 
{
    {
      std::lock_guard<std::mutex> lock(mutex);

      cancel_work = 1;
    }
    {
      std::lock_guard<std::mutex> lock(mutex);
      writ = true;
      cv.notify_all();
    }
}


// Thread 2 executes this function
void thread2_func() 
{
    std::unique_lock<std::mutex> lock(mutex);

    cv.wait(lock, []{ return writ; } );

    int tmp = cancel_work;
    std::cout << tmp; // will print 1
}

int main()
{
   std::thread t1(thread1_func);
   std::thread t2(thread2_func);

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

   return 0;
}

and now thread2_func happens after thread1_func and all is good. The read is guaranteed to be 1.