Question

我想创建解析器，它将解析以alpha或_开头的标识符，并且它可能在正文中包含alpha，num或_

这是我到目前为止所做的：

#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/spirit/home/support/iterators/line_pos_iterator.hpp>
#include <boost/spirit/repository/include/qi_confix.hpp>
#include <boost/spirit/include/phoenix_fusion.hpp>
#include <boost/spirit/include/phoenix_stl.hpp>

using namespace boost::spirit;

#include <boost/fusion/include/adapt_struct.hpp>

////////////////////////////////
// extra facilities
struct get_line_f
{
    template <typename> struct result { typedef size_t type; };
    template <typename It> size_t operator()(It const& pos_iter) const
    {
        return get_line(pos_iter);
    }
};

struct Position
{
    Position()
        : line(-1)
    {
    }

    size_t line;
};

struct Identifier : public Position
{
    Identifier()
        : Position()
        , name()
    {
    }

    std::string name;
};

BOOST_FUSION_ADAPT_STRUCT(Identifier,
                            (std::string, name)
                            (size_t,      line)
                          )

//
////////////////////////////////

template <typename Iterator>
struct source_identifier: qi::grammar<Iterator, Identifier(), qi::space_type>
{
    source_identifier() : source_identifier::base_type(start)
    {
        using qi::alpha;
        using qi::alnum;
        using qi::raw;
        using qi::_val;
        using qi::_1;

        namespace phx = boost::phoenix;
        using phx::at_c;
        using phx::begin;

        name %=     (qi::alpha | "_")
                >> *(qi::alnum | "_");

        start = raw [ name[at_c<0>(_val) = _1] ]
                    [
                        at_c<1>(_val) = get_line_(begin(_1))
                    ]
        ;
    }

    boost::phoenix::function<get_line_f> get_line_;
    qi::rule<Iterator, Identifier(), qi::space_type> start;
    qi::rule<Iterator, std::string()> name;
};

如果标识符中有“_”，则返回空名称的原因，如果没有

，则会起作用

Answer 1

"_"与qi::lit("_")相同，都与字符'_'匹配，但不合成任何属性。你想要的是qi::char_('_')。您可以找到更多信息here。

提升精神解析标识符

1 个答案: