c++ - 解析为类(不是结构)

Question

下面我展示了一个编辑过的 spirits employee 示例，它没有编译。我要解决的问题是解析为类而不是结构。我知道，除了公共(public)/私有(private)之外，它完全相同。但是在将类/结构存储到 vector 中之前，我需要有一个构造函数来工作。如何更改 BOOST_FUSION_ADAPT_STRUCT？

我怎样才能让它运行？

// 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;
}

Answer 1

你可以

1。适应为ADT

通过添加 getter/setter。另一个答案链接到这个作为解决方案:http://www.boost.org/doc/libs/release/libs/fusion/doc/html/fusion/adapted/adapt_adt.html

2。使用语义 Action :

查看 Live On Coliru

变化:

1. 使类默认可构造

2. 从规则的语义操作中显式调用构造函数:

       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<<所以你仍然可以打印类进行调试

3。提供定制点

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