Am Thursday 12 November 2009 15:00:09 schrieb Cosimo Calabrese:
namespace std { template <typename Graph1, typename Graph2> std::ostream& operator<<(std::ostream& os, const typename boost::graph_traits<boost::union_graph<Graph1, Graph2>
>::vertex_descriptor& v)
{ if ( boost::get<WR1<graph_traits<Graph1>::vertex_descriptor> >( &v ) ) os << boost::get<WR1<graph_traits<Graph1>::vertex_descriptor>
(v)._data;
else os << boost::get<WR2<graph_traits<Graph2>::vertex_descriptor> >(v)._data; return os; } }
the template parameters (Graph1/2) must be deducable from the function parameter (v). this is not possible in this example. consider the following, simpler, case: template<class T> struct A{ typedef T type; }; template<class A_T> void func(typename A_T::type); int main(){ int a; func<A<int> >(); //ok func(a); //error } the compiler has no way of knowing that foo is supposed to be instantiated with A_T == A, unless you provide an explicit template argument (which you don't when you call an operator).