[Boost-users] [Boost.Thread] unique_future without && ?

Subject: [Boost-users] [Boost.Thread] unique_future without && ?
From: John M. Dlugosz (mpbecey7gu_at_[hidden])
Date: 2012-04-06 03:20:09

My code needs to be compatible with a platform that doesn't have rvalue references 
available, for the time being.
I've avoided explicit uses of move and made my && arguments fall back to const&.
But I'm not sure about returning a unique_future.  I see that unique_future<T> has a 
special move-the-guts type, with an implicit conversion operator and a constructor.  But 
do I need to do anything special to declare a function returning one, and likewise the 
return statement itself?  The code in packaged_task ( unique_future<R> 
packaged_task<R>::get_future() ) is not conditionally compiled based on 
BOOST_NO_RVALUE_REFERENCES, so I'm supposing that it just works by itself.  Is that correct?

But maybe it only works by itself in some cases?  In that code, the return statement names 
a constructor directly.
                 return unique_future<R>(task);

In my code, I'm returning a unique_future that I already made somewhere else (actually, 
the result of another function call) so it would be wanting to call the copy constructor, 
I would think.

So if I write
     unique_future<int> foo();
     unique_future<int> bar() { return foo(); }

what _should_ happen is that the existing value uses a user-defined conversion to 
construct the "copy", right?  So I should expect this to work correctly if the thing I'm 
returning is already exactly the right type, as well as when naming the constructor directly.

—John