如何使用 boost::tuple 作为 boost::spirit 规则中的属性

我在boost::spirit有以下规则：

typedef boost::tuple<int, int> Entry;
qi::rule<Iterator, Entry(), Skipper> entry;
entry = qi::int_ >> qi::int_;

但是第二个int不会写入元组。有没有办法让它在不必使用boost::fusion::tuple的情况下工作？

如果我使用std::pair它就可以工作，那么为什么我不能使用boost::tuple呢？

下面是一个完整的编译示例：

#include <iostream>
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/include/tuple.hpp>
#include <boost/tuple/tuple.hpp>
namespace qi = boost::spirit::qi;
// works:
// #include <boost/fusion/include/std_pair.hpp>
// typedef std::pair<int, int> Entry;
// doesn't work:
typedef boost::tuple<int, int> Entry;
template <typename Iterator, typename Skipper>
struct MyGrammar : qi::grammar<Iterator, Entry(), Skipper> {
  MyGrammar() : MyGrammar::base_type(entry) {
    entry = qi::int_ >> qi::int_;
  }
  qi::rule<Iterator, Entry(), Skipper> entry;
};
int main() {
  const std::string in = "1 3";
  typedef std::string::const_iterator It;
  It it = in.begin();
  Entry entry;
  MyGrammar<It, qi::space_type> gr;
  if (qi::phrase_parse(it, in.end(), gr, qi::space, entry)
      && it == in.end()) {
    std::cout << boost::get<0>(entry) << "," << boost::get<1>(entry) << std::endl;
  }
  return 0;
}