Question

我有一个代码库，它与下面的代码完全相同。我尝试生成一个带有两倍变量内容的文本文件。我觉得答案是在语义动作和_a和_val中，但即使使用文档也无法通过。

你将如何做到：＆＃34; TOTO＆＃34;在str 并输出：托托的一些东西

即如何在业力中重用已解析的变量？

struct data
 {
  std::string str;
 };

BOOST_FUSION_ADAPT_STRUCT(
                      data,
                      (std::string, str)
                      )

template <typename Iterator>
struct data: karma::grammar<Iterator, data() >
{
    data():data::base_type(start)
    {
        start = karma::string << karma::lit("some stuff") << karma::string; //Second string is in fact the first one 
    }
    karma::rule<Iterator, data()> start;
};

解决方案（根据以下帖子：）

#include <iostream>
#include <string>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/spirit/include/karma.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_fusion.hpp>
#include <boost/spirit/include/phoenix_stl.hpp>
#include <boost/spirit/include/phoenix_object.hpp>
#include <boost/spirit/include/support_istream_iterator.hpp>
#include <boost/spirit/include/support_iso8859_1.hpp>

namespace ast
{
    struct data
    {
        std::string str;
    };
}

BOOST_FUSION_ADAPT_STRUCT(
                          ast::data,
                          (std::string, str)
                          )

namespace karma = boost::spirit::karma;

namespace parser
{

    template <typename Iterator>
    struct data: karma::grammar<Iterator, ast::data() >
    {
        data():data::base_type(start)
        {
            start =
                karma::string[karma::_1 =  boost::phoenix::at_c<0>(karma::_val)] <<
                karma::lit("some stuff") <<
                karma::string[karma::_1 =  boost::phoenix::at_c<0>(karma::_val)]
                ;
        }
        karma::rule<Iterator, ast::data()> start;
    };
}

main()
{
    ast::data d;
    d.str = "toto";
    std::string generated;
    typedef std::back_insert_iterator<std::string> iterator_type;
    parser::data<iterator_type> d_p;
    iterator_type sink(generated);
    karma::generate(sink, d_p, d);
    std::cout << generated << std::endl;
}

Answer 1

这应该可以解决问题：

start = karma::string[karma::_1 = karma::_val] 
    << karma::lit("some stuff") 
    << karma::string[karma::_1 = karma::_val];

使用boost karma重用已解析的变量

1 个答案: