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我有一条规则应该返回一个 Boost.Fusion ASSOC_STRUCT。我正在尝试将规则解析器解析的结果分配给 _val，但我无法使其工作。我将跳过谈话，直接向您展示相关代码。

#include <boost/fusion/include/define_assoc_struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/fusion/include/as_vector.hpp>
#include <boost/spirit/include/qi_symbols.hpp>
#include <iostream>
#include <string>

namespace keys {
   struct actor_key;
   struct action_key;
   struct money_key;
 }

 BOOST_FUSION_DEFINE_ASSOC_STRUCT(
    (), ActionLine,
    (int, action, keys::action_key)
    (int, amount, keys::money_key)
    (int, actor, keys::actor_key)
  )

int main ()
{
  namespace qi = boost::spirit::qi;
  using qi::_1;
  using qi::_2;
  using qi::_a;
  using qi::int_;
  using qi::lit;
  using boost::spirit::_val;
  using boost::fusion::as_vector;

  qi::symbols<char, int> name_parser_;
  name_parser_.add("name4", 4);

  std::string input("string1 ($7) string2 name4");
  std::string::const_iterator f(input.begin()), l(input.end());  
  ActionLine expected{0, 7, 4}, result;

  //The following rule is supposed to parse input of the form
  //"string1 ($[integer]) string2 [name]"
  //and return a triple (ActionLine) with the values { 0, value of (int_), value of (name_parser_) }
  qi::rule<std::string::const_iterator, ActionLine()> action_line_ =
     lit("string1 ($") >> int_ >> lit(") string2 ") >> name_parser_
     // [ _val = ActionLine{0, _1, _2} ]; // this is what I am trying to achieve
    [ _val = ActionLine{0, 7, 4} ]; //this compiles, but of course is not what I need

   bool b =
     qi::parse(f, l, action_line_, result) &&
     as_vector(result) == as_vector(expected);

   std::cout << "test: " << std::boolalpha << b << std::endl;
   return 0;
 }

（使用 g++ above_file.cpp -std=c++0x 编译）我的实际应用程序中的编译器错误与此示例中的有些不同，但类似于（在 _val = ActionLine{0, _1, _2} 的行中） : 没有调用::ActionLine::ActionLine()的匹配函数，我猜它不能将_1和_2转换为int。

我还尝试添加本地 int 变量并使用它们来复制解析的值，但它不起作用，没有使用 boost::phoenix::at(_1,0), boost::phoenix::at(_1,1 ）（我发现这些想法在这里提升了精神语义动作参数）

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1 回答 1

2

你需要phoenix::construct在你的语义动作中使用来做你想做的事。

#include <boost/fusion/include/define_assoc_struct.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> //CHANGE:you need to include this to use "phoenix::construct"
#include <boost/fusion/include/as_vector.hpp>
#include <boost/spirit/include/qi_symbols.hpp>
#include <iostream>
#include <string>

namespace keys {
   struct actor_key;
   struct action_key;
   struct money_key;
 }

 BOOST_FUSION_DEFINE_ASSOC_STRUCT(
    (), ActionLine,
    (int, action, keys::action_key)
    (int, amount, keys::money_key)
    (int, actor, keys::actor_key)
  )

int main ()
{
  namespace qi = boost::spirit::qi;
  namespace phx = boost::phoenix;
  using qi::_1;
  using qi::_2;
  using qi::_a;
  using qi::int_;
  using qi::lit;
  using boost::spirit::_val;
  using boost::fusion::as_vector;

  qi::symbols<char, int> name_parser_;
  name_parser_.add("name4", 4);

  std::string input("string1 ($7) string2 name4");
  std::string::const_iterator f(input.begin()), l(input.end());  
  ActionLine expected{0, 7, 4}, result;

  //The following rule is supposed to parse input of the form
  //"string1 ($[integer]) string2 [name]"
  //and return a triple (ActionLine) with the values { 0, value of (int_), value of (name_parser_) }
  qi::rule<std::string::const_iterator, ActionLine()> action_line_ =
     (lit("string1 ($") >> int_ >> lit(") string2 ") >> name_parser_) //CHANGE:the parentheses are important otherwise the semantic action would be attached to name_parser_
    [ _val = phx::construct<ActionLine>(0, _1, _2) ]; //CHANGE: you need to use phoenix to use the placeholders _1, _2, etc

   bool b =
     qi::parse(f, l, action_line_, result) &&
     as_vector(result) == as_vector(expected);

   std::cout << "test: " << std::boolalpha << b << std::endl;
   std::cout << at_key<keys::action_key>(result)<< ", " << at_key<keys::money_key>(result)<< ", " << at_key<keys::actor_key>(result) << std::endl;

   return 0;
 }
于 2013-01-18T11:03:05.540 回答