返回右值的shared_ptr成员

Question

In C++ Concurrency In Action - Practical MultiThreading page 167, there's the code snipet

 std::shared_ptr<T> wait_and_pop()
 {
     std::unique_ptr<node> const old_head=wait_pop_head();
     return old_head->data;
 }

Why do we have to move assign rvalue wait_pop_head() to a const variable first? Is there a reason why we can't shorthand the code to following?

std::shared_ptr<T> wait_and_pop()
{
    return wait_pop_head()->data;
}

Answer 1

Indeed, there is no reason that your alternative could not be used.

The temporary will live long enough.

But some people just prefer to write it out.