Question

In one of my rules a have qi::locals<> with stl container as parameter. When I compile my program without BOOST_SPIRIT_QI_DEBUG, everything works fine. But when I turn BOOST_SPIRIT_QI_DEBUG on, program can’t be compiled because of missing operator <<.

Simplified snippet:

typedef std::list< int > TlstValues;

template <typename Iterator, typename Skipper>
struct G_test : qi::grammar<Iterator, Skipper>
{
  ...
  G_test() : G_test::base_type(rule_test)
  {
    ...
    BOOST_SPIRIT_DEBUG_NODE(rule_test);
  }

  qi::rule<Iterator, qi::locals<TlstValues>, Skipper> rule_test;
}

error:

error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'TlstExpressionItems' (or there is no acceptable conversion)

I tried to define streaming operator for TlstValues but without effect. Error was still the same. My operator definition:

std::ostream& operator << (std::ostream& os, TlstValues & z)
{
return os;
} 

Can someone please advise me how to define required operator to make BOOST_SPIRIT_DEBUG work?

Thanks and regards
Rick

Answer 1

The simplest way to force ADL to find your streaming operator is to put it into the namespace std:

namespace std
{
    std::ostream& operator<< (std::ostream& os, TlstValues& z)
    {
        // do your stuff here
        return os;
    } 
}

Yes, I know, this is formally not allowed. However, from a pragmatic standpoint it’s still the easiest way.

The Standards conformant solution is to specialize the Spirit customization point print_attribute_debug for your attribute type:

// your specialization needs to be in namespace boost::spirit::traits
namespace boost { namespace spirit { namespace traits
{
    template <typename Out>
    struct print_attribute_debug<Out, TlstValues>
    {
        static void call(Out& out, TlstValues const& val)
        {
            // do your output here; Out is a std::ostream
        }
    };
}}}