on Fri Jul 27 2007, "JOAQUIN LOPEZ MU?Z" <joaquin-AT-tid.es> wrote:
According to the MPL reference, a placeholder expression X<a1,..,an> must satisfy (among others) the following condition:
"All of X's template parameters, including the default ones, are types."
According to that the following should then compile without errors:
#include <boost/mpl/apply.hpp> #include <boost/mpl/assert.hpp> #include <boost/type_traits/is_same.hpp>
using namespace boost; using namespace boost::mpl;
template<typename T,int=0> struct foo { template<typename Q> struct apply{ typedef Q type; }; };
typedef apply1<foo<_1>,int>::type t1;
BOOST_MPL_ASSERT((is_same<t1,int>));
int main(){}
I don't see how you can conclude the above will compile without error just from (a part of) the definition of placeholder expression. Are you saying that since foo<_1> is not a placeholder expression, when applied to int it should be treated as a metafunction class? -- Dave Abrahams Boost Consulting http://www.boost-consulting.com The Astoria Seminar ==> http://www.astoriaseminar.com