On Thu, Jan 12, 2012 at 2:27 AM, Luke J. West <ljw@oopstart.com> wrote:
Hi. I'm trying to implement a template function that 'moves' it's parameter. Here's a simple moveable class...
struct my_class { my_class() {} my_class(BOOST_RV_REF(my_**class)) {}
my_class& operator=(BOOST_COPY_ASSIGN_**REF(my_class) c) {return *this;}
BOOST_COPYABLE_AND_MOVABLE(my_**class); };
And here is the function...
template<typename T > void steal(BOOST_COPY_ASSIGN_REF(T) c) {}
Unrelated: Why use BOOST_COPY_ASSIGN_REF( T ) as the function parameter? Pretty sure this macro is only to be used within a operator= declaration/definition. Should probably use BOOST_RV_REF( T ).
so it works fine when I do this...
int main() { my_class c; steal<my_class >(c); }
but not when I do this...
int main() { int c; steal<int >(c); }
It's as if 'int' is not a movable type which of course it trivially is. I read that Boost.Container classes are 'move aware' and I assume boost::vector<int > works, so what am I doing wrong - how to write my 'steal' function capable of moving it's argument when it's moveable class, but not breaking for primitives?
Thanks,
Luke (hopeful newbie!)
Hmmm...right...I think the compiler doesn't like instantiating boost::rv<int>. AFAIK a trac ticket had just been recently filed to this effect. Note that in C++11, lvalues are not implicitly convertible to rvalue references. So if you actually wanted steal to take an (emulated) rvalue reference, you should still call it with an enclosed boost::move: steal(boost::move(c)) and if you know that steal will always be accepting a moveable object, you can declare it as template< class T > void steal(BOOST_RV_REF( T )); Can you give a better idea of what you're trying to achieve? - Jeff