On 16/07/2012 02:11, Kjell Ahlstedt wrote: >> >> > Thanks. > I haven't tested it (and will probably not do it) but it seems quite > reasonable. > I suppose it will also require some calls to BOOST_TYPEOF_REGISTER_TEMPLATE.
No it doesn't. Here's my full test-case:
//-------------------------------
#include <iostream>
#include <boost/type_traits.hpp>
#include <sigc++/sigc++.h>
namespace sigc
{
template <typename Functor>
struct functor_trait<Functor, false>
{
typedef BOOST_TYPEOF_TPL (sigc::mem_fun
(boost::declval<Functor&> (),
&Functor::operator())) _intermediate;
typedef typename _intermediate::result_type result_type;
typedef Functor functor_type;
};
}
struct Foo
{
bool operator() ()
{
std::cout << "Hello world!" << std::endl;
}
};
int main ()
{
sigc::slot<bool> slot = Foo ();
slot ();
return 0;
}
//-------------------------------
--
Kind regards,
Loong Jin
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