------- Comment #4 from dgregor at gcc dot gnu dot org 2008-09-24 20:20 -------
GCC is doing the right thing here. In this constructor:
Thing2(Thing2&& o) : Thing(o) { }
the parameter "o" is treated as an lvalue, because it has a name. Using
std::move(o) to treat it as an rvalue.
Similarly, there are no automatically generated move constructors or
move-assignment operators, so if you leave Thing2() empty, you'll just get the
copy constructor and therefore do a copy.
--
dgregor at gcc dot gnu dot org changed:
What |Removed |Added
----------------------------------------------------------------------------
Status|UNCONFIRMED |RESOLVED
Resolution| |INVALID
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=36461
