Return std::move(unique_lock<>)

Question

When I read the book "C++ in concurrent",I see the code about return the std::move(unique_lock<>), for example:

    std::unique_lock<std::mutex> wait_for_data()
    {
        std::unique_lock<std::mutex> head_lock(head_mutex);
        data_cond.wait(head_lock,[&]{return head!=get_tail();});
        return std::move(head_lock);
    }

    std::unique_ptr<node> do()
    {
        std::unique_lock<std::mutex> head_lock(wait_for_data());
        return pop_head();
    }

Are there two locks and two unlocks? Is it same with next example which just have one lock and unlock ? What's the meaning about return the unique_lock<>?

    std::unique_ptr<node> do()
    {
        std::unique_lock<std::mutex> head_lock(head_mutex);
        data_cond.wait(head_lock,[&]{return head!=get_tail();});
        return pop_head();
    }

Thanks in advance for any help!!

Answer 1

In wait_for_data() unique_lock constructed from mutex (head_mutex). Then in do() unique_lock constructed from unique_lock rvalue that wait_for_data() returned. So in do() you actually use the same unique_lock that wait_for_data() created.