解析成类(不是结构)

时间:2014-10-23 12:16:38

标签: c++ boost-spirit boost-fusion

下面我展示了一个不编译的编辑灵魂员工示例。我想解决的问题是解析为不是结构的类。我知道,除了公共/私人之外,它是完全相同的。但是我需要在将类/结构存储到向量之前使用构造函数。如何更改BOOST_FUSION_ADAPT_STRUCT?

如何让它运行?

// STD HEADER
#include <iostream>
#include <string>
#include <complex>

// BOOST HEADER
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_object.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/include/io.hpp>

namespace client
{
    namespace qi = boost::spirit::qi;

    namespace ascii = boost::spirit::ascii;

    class employee
    {
    public:
        employee (
            int _age
          , std::string _surname
          , std::string _forename
          , double _salary
        );

    private:
        int age_;
        std::string surname_;
        std::string forename_;
        double salary_;
    };

    employee::employee (
        int _age
      , std::string _surname
      , std::string _forename
      , double _salary
    ) : age_(_age)
      , surnemae_(_surename)
      , forename_(_forename)
      , double salary_
    {
        // do some important stuff
    }

}

// WHAT TO DO HERE?
BOOST_FUSION_ADAPT_STRUCT(
    client::employee
)

namespace client
{
    template <typename Iterator>
    struct employee_parser
      : qi::grammar<Iterator, employee(), ascii::space_type>
    {
        employee_parser() : employee_parser::base_type(start)
        {
            using qi::int_;
            using qi::lit;
            using qi::double_;
            using qi::lexeme;
            using ascii::char_;

            quoted_string %= lexeme['"' >> +(char_ - '"') >> '"'];

            start %=
                lit("employee")
                >> '{'
                >>  int_ >> ','
                >>  quoted_string >> ','
                >>  quoted_string >> ','
                >>  double_
                >>  '}'
                ;
        }

        qi::rule<Iterator, std::string(), ascii::space_type> quoted_string;
        qi::rule<Iterator, employee(), ascii::space_type> start;
    };
}

int main()
{
    std::cout << "/////////////////////////////////////////////////////////\n\n";
    std::cout << "\t\tAn employee parser for Spirit...\n\n";
    std::cout << "/////////////////////////////////////////////////////////\n\n";

    std::cout
        << "Give me an employee of the form :"
        << "employee{age, \"surname\", \"forename\", salary } \n";
    std::cout << "Type [q or Q] to quit\n\n";

    using boost::spirit::ascii::space;
    typedef std::string::const_iterator iterator_type;
    typedef client::employee_parser<iterator_type> employee_parser;

    employee_parser g;
    std::string str;
    while (getline(std::cin, str))
    {
        if (str.empty() || str[0] == 'q' || str[0] == 'Q')
            break;

        client::employee emp;
        std::string::const_iterator iter = str.begin();
        std::string::const_iterator end = str.end();
        bool r = phrase_parse(iter, end, g, space, emp);

        if (r && iter == end)
        {
            std::cout << boost::fusion::tuple_open('[');
            std::cout << boost::fusion::tuple_close(']');
            std::cout << boost::fusion::tuple_delimiter(", ");

            std::cout << "-------------------------\n";
            std::cout << "Parsing succeeded\n";
            std::cout << "got: " << boost::fusion::as_vector(emp) << std::endl;
            std::cout << "\n-------------------------\n";
        }
        else
        {
            std::cout << "-------------------------\n";
            std::cout << "Parsing failed\n";
            std::cout << "-------------------------\n";
        }
    }

    std::cout << "Bye... :-) \n\n";
    return 0;
}

2 个答案:

答案 0 :(得分:5)

你可以

1。适应ADT

添加getter / setter。另一个答案与此相关联:http://www.boost.org/doc/libs/release/libs/fusion/doc/html/fusion/adapted/adapt_adt.html

2。使用语义操作:

查看 Live On Coliru

的变化:

  1. 使类默认可构造
  2. 从规则的语义操作中明确调用构造函数:

        start =
            lit("employee")
            >> ('{'
            >>  int_ >> ','
            >>  quoted_string >> ','
            >>  quoted_string >> ','
            >>  double_
            >>  '}'
            ) [ qi::_val = boost::phoenix::construct<employee>(qi::_1, qi::_2, qi::_3, qi::_4) ]
            ;
    
  3. 可选择重载operator<<,这样您仍然可以打印类进行调试

  4. 3。提供自定义点

    请参阅http://www.boost.org/doc/libs/1_56_0/libs/spirit/doc/html/spirit/advanced/customize/assign_to.html

答案 1 :(得分:1)

Boost.Fusion 提供了宏BOOST_FUSION_ADAPT_STRUCT_ADT,它为您的用例使用了setter和getter函数。有关详细信息,请参阅:http://www.boost.org/doc/libs/1_50_0/libs/fusion/doc/html/fusion/adapted/adapt_adt.html