// This is a silly example, but it demonstrates that the type of a lazy default
// probably should not be evaluated when the parameter is given explicitly
// (since the type of the default is never needed in that case).

#include <boost/parameter.hpp>

BOOST_PARAMETER_NAME(param);

template <typename T> struct empty {};

struct my_default_computer {
#if 0
  // What I need to do now (broken):
  typedef empty<int>::type result_type;
#else
  // What I want to do:
  template <typename T>
  struct result {typedef typename empty<T>::type type;};
#endif
};

int main(int, char**) {
  (_param = 0)[_param || my_default_computer()];
  return 0;
}