检测 Spirit 语义动作中的参数类型

Sometimes, the compiler will complain about assignment or type incompatibilities and I have no clue what's wrong. The problem occurs in two main areas:

• 预测规则/表达式的综合属性类型
  • 因此，预测哪些类型的属性可以合法地定义为规则的公开属性(依赖于内置转换、融合适配器或 Spirit 自定义点)
  • 编译器将能够编译函数调用
  • 调用不会在过程中调用不必要的隐式转换

  编译器错误并不完全易于处理，要么文档有误，要么我误解了它.

  The compiler error is not exactly tractable, and either the documentation is wrong, or I misunderstood it.

  无论如何，有没有办法确切地找出精神传递到我的语义动作中的内容?

  Is there a way to find out exactly what Spirit passes into my semantic action, anyway?

  struct mybase             { int a,b; };
  struct myderived : mybase { int c,d; };
  
  BOOST_FUSION_ADAPT_STRUCT(mybase,    (int,a)(int,b));
  BOOST_FUSION_ADAPT_STRUCT(myderived, (int,a)(int,b)(int,c)(int,d));
  
  auto base_expr = int_ >> int_; // avoids assigning to struct attribute
  
  rule<decltype(f), mybase()   , space_type> base_       = int_ >> int_;
  rule<decltype(f), myderived(), space_type> derived_    = base_ >> int_ >> int_;
  
  myderived data;
  bool ok = phrase_parse(f,l,derived_,space,data);
  

  此代码无法编译，存在大量难以理解的错误.

  This code won't compile, with a huge amount of impenetrable errors.

  (粗略改编自 精神一般列表)

  推荐答案

  为了清楚起见 - 这里的错误是 base_ >>内部_>>int_ 被用作创建 myderived 的规则的表达式，并且由于 base_ 固定为类型 mybase，我们d 必须从 mybase 和两个 int 创建一个 myderrived，但是没有什么可以告诉 Spirit 如何做到这一点.

  For clarity - the error here is that base_ >> int_ >> int_ was used as the expression for a rule that creates a myderived, and since base_ is fixed to type mybase, we'd have to create a myderrived from a mybase and two ints, but there's nothing to tell Spirit how to do that.

  您可以使用 boost 打印出 boost 通过解析 base_ >> 创建的值的类型.内部_>>int_ 通过定义一个可以接受任何参数的函子，并告诉您它们是什么(以下代码改编自 SO chat 上的一些代码):

  You can get boost to print out the type of the value that boost creates from parsing base_ >> int_ >> int_ by defining a functor that will take any parameters, and tell you what they are (the following code is adapted from some code sehe put on SO chat):

  struct what_is_the_attr
  {
      template <typename> struct result { typedef bool type; };
  
      template <typename T>
      static void print_the_type()
      {
          std::cout << "    ";
          std::cout << typeid(T).name();
          if(std::is_const<typename std::remove_reference<T>::type>::value)
              std::cout << " const";
          if(std::is_rvalue_reference<T>::value)
              std::cout << " &&";
          else if(std::is_lvalue_reference<T>::value)
              std::cout << " &";
      }
  
      template <typename Th, typename Th2, typename... Tt>
      static void print_the_type()
      {
          print_the_type<Th>();
          std::cout << ",
  ";
          print_the_type<Th2, Tt...>();
      }
  
      template <typename... Ts>
      void operator()(Ts&&...) const
      {
          std::cout << "what_is_the_attr(
  ";
          print_the_type<Ts...>();
          std::cout << ")" << std::endl;
      }
  };
  

  然后要使用它，请在初始化程序的语义操作中为您的错误规则使用上述角色:

  Then to use it, use the above actor in a semantic action on initializer for your faulty rule:

  std::string input = "1 2 3 4";
  auto f(std::begin(input)), l(std::end(input));
  
  rule<decltype(f), mybase()   , space_type> base_    = int_ >> int_;
  rule<decltype(f), myderived(), space_type> derived_ = (base_ >> int_ >> int_)[what_is_the_attr()];
  
  myderived data;
  bool ok = phrase_parse(f,l,derived_,space,data);
  

  注意，您不能通过 %= 使用自动属性传播(除非您从规则的声明类型中删除了暴露的属性类型).

  Note, you cannot use automatic attribute propagation with %= (unless you remove the exposed attribute type from the rule's declared type).

  运行它应该会产生一个编码类型，可以用 c++filt -t 解码:生活在 Coliru 上

  Running this should then yield an encoded type, which can be decoded with c++filt -t: Live On Coliru

  $ g++ 9404189.cpp -std=c++0x
  $ ./a.out |c++filt -t
  what_is_the_attr(
      boost::fusion::vector3<mybase, int, int> &,
      boost::spirit::context<boost::fusion::cons<boost::spirit::unused_type&, boost::fusion::nil>, boost::fusion::vector0<void> > &,
      bool &)
  

  第一行，boost::fusion::vector3<mybase, int, int>，至少告诉你 boost 正在尝试从 3 个 mybase 类型的对象创建你的返回类型、int 和 int.

  The first line, boost::fusion::vector3<mybase, int, int>, least tells you that boost is trying to create your return type from 3 objects of types mybase, int and int.