Since we are passing the object by reference to std::async, it will call the operator () on same object, then why its member variable is not getting updated

struct Y
{
    int m_val;
    Y():m_val(0){}
    double operator()(double val)
    {
        m_val = val*val;
        return m_val;
    }
};
int main()
{
    Y y;
    auto f=std::async(std::ref(y),2);
    cout <<"Returned value " << f.get() << ", Modified value " << y.m_val << endl;
    return 0;
}

output-

Returned value 4, Modified value 0

As per my understanding it should have called y(2) so y.m_val should be updated to 4, however it is printed as 0 in output. Please clarify what am I missing.

CodePudding user response：

Before C 17, the evaluation of arguments to << was unsequenced. That means you have no guarantee that f.get() would be called before y.m_val's value is taken.

As a consequence, your program has a potential data race, and therefore undefined behavior.

Since C 17, the evaluation order is specified as left-to-right, and so the result you expect is guaranteed.

For C 11/14 you can fix this by separating the output statement into multiple statements:

std::cout << "Returned value " << f.get();
std::cout << ", Modified value " << y.m_val << endl;

or even by calling f.get() in a separate statement before the output (storing the result to a variable), which is generally a good idea anyway if your call has side effects.