4

下面的代码（改编自 Spirit qi mini_xml 示例）无法编译。存在与brac具有递归属性的规则相关的错误boost::variant。
但是，所有注释掉的bracdo compile 版本。

我很想知道是什么让简单的字符串解析器在这种情况下如此特别：

#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;
}

部分错误信息：

/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。

4

1 回答 1

5

我建议问题是属性兼容性。与文档相反， ascii::string 解析器似乎公开了一个迭代器范围而不是一个字符串。

name = string("no way");

没问题，因为 ascii::string 暴露的属性可以毫无困难地强制转换为规则的属性类型。

但是，brac规则的属性类型是ast_branch，它只是一个变体，它具有其中一种可能包含的类型。因此，该ast_branch类型有几个构造函数，Spirit 并不清楚哪个构造函数适合这种特定的转换。

有几种方法（除了你已经展示的方法）：

  • 利用attr_cast

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

  • 利用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);
            }
        };
}}}

每一个都有相同的效果：让Spirit实现使用什么属性转换。

这是一个完整的工作示例，显示了所有三个：

// #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;
}
于 2012-07-10T22:04:10.623 回答