c – 带有boost变量递归包装器的字符串解析器

下面的代码(改编自spirit qi mini_xml示例)无法编译.存在与规则brac相关的错误,该错误具有递归boost :: variant的属性.
但是,所有注释掉的brac版本都会编译.

我非常好奇在这种情况下知道是什么让简单的字符串解析器如此特殊：

#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_fusion.hpp>
#include <boost/spirit/include/phoenix_stl.hpp>
#include <boost/spirit/include/phoenix_object.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/variant/recursive_variant.hpp>

#include <string>
#include <vector>

namespace client
{
   namespace fusion = boost::fusion;
   namespace phoenix = boost::phoenix;
   namespace qi = boost::spirit::qi;
   namespace ascii = boost::spirit::ascii;

   struct ast_node;

   typedef boost::variant<
      boost::recursive_wrapper<ast_node>,
      std::string
   > ast_branch;

   struct ast_node
   {
      std::string text;
      std::vector<ast_branch> children;
   };
}

BOOST_FUSION_ADAPT_STRUCT(
      client::ast_node,
      (std::string, text)
      (std::vector<client::ast_branch>, children)
)

namespace client
{
   template <typename Iterator>
      struct ast_node_grammar
      : qi::grammar<Iterator, ast_branch(), ascii::space_type>
      {
         ast_node_grammar()
            : ast_node_grammar::base_type(brac)
         {
            using qi::_1;
            using qi::_val;
            using ascii::char_;
            using ascii::string;

            name %= *char_;

            brac %= string("no way") ;
//            brac = string("works")[_val = _1] ;
//            brac %= string("this") | string("works");
//            brac %= name ; // works
//            brac %= *char_ ; // works
         }
         qi::rule<Iterator, std::string()> name;
         qi::rule<Iterator, ast_branch(), ascii::space_type> brac;
      };
}


int main(int argc, char **argv)
{
   typedef client::ast_node_grammar<std::string::const_iterator> ast_node_grammar;
   ast_node_grammar gram;
   client::ast_branch ast;

   std::string text("dummy");
   using boost::spirit::ascii::space;
   std::string::const_iterator iter = text.begin();
   std::string::const_iterator end = text.end();
   bool r = phrase_parse(iter, end, gram, space, ast);
   return r ? 0 : 1;
}

部分错误消息：

/usr/include/boost/spirit/home/qi/detail/assign_to.hpp:38:17: error: No match for ‘boost::variant<
        boost::recursive_wrapper<client::ast_node>, basic_string<char> 
>::variant(
        const __normal_iterator<const char *, basic_string<char> > &, const __normal_iterator<
            const char *, basic_string<char> > &)’

提前致谢.

解决方法:

我建议问题是属性兼容性.与the documentation相反,ascii :: string解析器似乎暴露了迭代器范围而不是字符串.

name = string("no way");

没问题,因为ascii :: string公开的属性可以毫不费力地强制进入规则的属性类型.

但是,brac规则的属性类型是ast_branch,它只是一个变体,其中包含一个可能包含的类型.因此,ast_branch类型有几个构造函数,并且Spirit不清楚哪个适合这种特定的转换.

有几种方法(除了你已经展示的方法)：

>使用attr_cast

brac = qi::attr_cast( string("no way") );

>使用as_string

brac = qi::as_string[ string("no way") ];

>使用自定义点

namespace boost { namespace spirit { namespace traits {
    template <typename It>
        struct assign_to_attribute_from_iterators<client::ast_branch, It>
        {
            static void call(It const& f, It const& l, client::ast_branch& val)
            {
                val = std::string(f, l);
            }
        };
}}}

这些中的每一个都具有相同的效果：使Spirit实现要使用的属性转换.

这是一个完整的工作样本,显示所有三个：

// #define BOOST_SPIRIT_ACTIONS_ALLOW_ATTR_COMPAT
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/adapted.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/variant/recursive_variant.hpp>

#include <string>
#include <vector>

namespace client
{
   namespace fusion  = boost::fusion;
   namespace phoenix = boost::phoenix;
   namespace qi      = boost::spirit::qi;
   namespace ascii   = boost::spirit::ascii;

   struct ast_node;

   typedef boost::variant<
      boost::recursive_wrapper<ast_node>,
      std::string
   > ast_branch;

   struct ast_node
   {
      std::string text;
      std::vector<ast_branch> children;
   };
}

namespace boost { namespace spirit { namespace traits {
    template <typename It>
        struct assign_to_attribute_from_iterators<client::ast_branch, It>
        {
            static void call(It const& f, It const& l, client::ast_branch& val)
            {
                val = std::string(f, l);
            }
        };
}}}

BOOST_FUSION_ADAPT_STRUCT(
      client::ast_node,
      (std::string, text)
      (std::vector<client::ast_branch>, children)
)

namespace client
{
    template <typename Iterator>
        struct ast_node_grammar : qi::grammar<Iterator, ast_branch(), ascii::space_type>
    {
        ast_node_grammar()
            : ast_node_grammar::base_type(brac)
        {
            using qi::_1;
            using qi::_val;
            using ascii::char_;
            using ascii::string;

            name %= *char_;

            brac = string("works");
            brac = string("works")[_val = _1] ;
            brac %= string("this") | string("works");
            brac %= name ; // works
            brac %= *char_ ; // works

            brac = qi::as_string[ string("no way") ];
            brac = qi::attr_cast( string("no way") );
        }
        qi::rule<Iterator, std::string()> name;
        qi::rule<Iterator, ast_branch(), ascii::space_type> brac;
    };
}


int main(int argc, char **argv)
{
   typedef client::ast_node_grammar<std::string::const_iterator> ast_node_grammar;
   ast_node_grammar gram;
   client::ast_branch ast;

   std::string text("dummy");
   using boost::spirit::ascii::space;
   std::string::const_iterator iter = text.begin();
   std::string::const_iterator end = text.end();
   bool r = phrase_parse(iter, end, gram, space, ast);
   return r ? 0 : 1;
}

标签：c,parsing,boost-spirit,boost-variant
来源： https://codeday.me/bug/20190902/1787413.html