Is it safe to remove the c++ volatile here?

Is it safe to remove volatile from the definition of m_flag here? If m_flag is not volatile, what would stop compilers from optimizing away this loop: while (!m_flag) m_cv.wait(lock); ? Does the standard (post-C++11) specify explictly that such loops can't be optimized away in such cases?

#include <mutex>
#include <condition_variable>
#include <future>
#include <iostream>
using namespace std;

class foofoo
{
    volatile bool m_flag;
    mutex m_mutex;
    condition_variable m_cv;

public:
    void DoWork()
    {
        m_flag = false;
        unique_lock<mutex> lock(m_mutex);
        auto junk = async(std::launch::async, [this]()
        {
            {
                unique_lock<mutex> lock(m_mutex);
                m_flag = true;
            }
            m_cv.notify_one();
        });
        while (!m_flag) m_cv.wait(lock);
        cout << "ququ" << endl;
    }
};

int main()
{
    foofoo f;
    f.DoWork();
}