on Fri Jan 09 2009, Anthony Williams <anthony.ajw-AT-gmail.com> wrote:
boost::move should return T.
Sorry for jumping in here, but I'm not sure that it should. Consider
movable m; boost::move(m);
In C++0x, move(m) is equivalent to static_cast<movable&&>(m), which just obtains a reference. If boost::move<T> returns a T then the code above will create a temporary which moves the data out of m, and then destroy that temporary, leaving m a hollow shell.
So you're saying, in other words, that "move(x)" really means "you have permission to move x" but the one I proposed would mean "move it, now." OK, good point. So what about this horrible little proposal? template <class T> struct rv<T> : T { private: rv(); ~rv(); rv(rv const&); void operator=(rv const&); }; template <class T> boost::enable_if<is_class<T>, rv<T>&> move(T& x) { return static_cast<rv<T>& >(x); } Does that solve any problems? Yeah, I know it's not theoretically portable, but it should be portable in practice. Especially when it comes to emulating language features, I care less and less about the letter of the law :-) -- Dave Abrahams BoostPro Computing http://www.boostpro.com