thread_local和std :: future对象-对象的生存期是多少？

Question

Here's my test code:

vector<int> const & foo(int const counter)
{
    thread_local static vector<int> v{counter, counter + 1, counter + 2};
    return v;
}

int main()
{
    using myFut = future<vector<int> const &>;

    vector<myFut> futures;
    for(int i{0}; i < 5; ++i)
    {
        futures.push_back(async(launch::async, &foo, i * 3));
    }

    for(myFut & fut : futures)
    {
        vector<int> v{fut.get()}; // or vector<int> const & v{fut.get()};
        cout << v.size() << endl; // 0, I expect 3
    }

    return 0;
}

When foo() returns a thread can be destroyed - together with a thread_local variable. But since I'm using a std::future the lifetime of the variable should be prolonged until the call to std::future::get() , right? But in my case the std::future returns an empty vector. So what are the rules?

Answer 1

But since I'm using a std::future the lifetime of the variable should be prolonged until the call to std::future::get(), right?

That's not the case. The thread used by std::async will call set_value() on the std::promise associated with the future, and then it's free to terminate. Therefore, your thread-local variable may well be destroyed before std::future::get() returns or even before you call it.