boost::spirit 算术公式解析器无法编译

Question

我正在尝试为填充抽象语法树的算术表达式编写一个精神解析器。如果我不尝试填充 AST，解析器将编译，但在当前版本中失败（出现一个 24K 错误）。我正在使用带有 -std=c++11 的 clang++ 3.5.0，并在 Ubuntu 14.4 上运行。

#include <string>
#include <vector>
#include <utility>

#include <boost/spirit/include/qi.hpp>

#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/adapted.hpp>
#include <boost/fusion/include/adapted.hpp>

#include <boost/variant/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>

using std::string;
using std::vector;
using std::pair;

using boost::spirit::qi::grammar;
using boost::spirit::qi::space_type;
using boost::spirit::qi::rule;

struct Term; // forward dec
typedef boost::recursive_wrapper<Term> RWTerm;
typedef pair<char, RWTerm> OpAndRWTerm;
typedef pair<RWTerm, vector<OpAndRWTerm> > Expr;
typedef boost::variant<Expr, double> Factor;
typedef pair<char, Factor> OpAndFactor;
struct Term : public pair<Factor, vector<OpAndFactor> >{};

template<typename It>
struct formula_parser : grammar<It, Expr(), space_type> {
  formula_parser() : formula_parser::base_type(expr_rule) {
    using boost::spirit::qi::double_;
    using boost::spirit::ascii::char_;

    factor_rule %= double_ | parenthesis_rule;
    parenthesis_rule %= '(' >> expr_rule >> ')';
    op_and_factor_rule %= char_("/*") >> factor_rule;
    term_rule %= factor_rule >> *op_and_factor_rule;
    op_and_term_rule %= char_("+-") >> term_rule;
    expr_rule %= term_rule >> *op_and_term_rule;
  }
  rule<It, OpAndRWTerm(), space_type> op_and_term_rule;
  rule<It, Expr(), space_type> expr_rule;
  rule<It, OpAndFactor(), space_type> op_and_factor_rule;
  rule<It, RWTerm(), space_type> term_rule;
  rule<It, Expr(), space_type> parenthesis_rule;
  rule<It, Factor(), space_type> factor_rule;
};

int main() {
  formula_parser<string::const_iterator> grammar;
}

我从错误消息中了解到，fusion 在规则 term_rule 中混淆了 Types Factor 和 RWTerm。

我做错了什么？

Answer 1

如果我改变两件事，它会为我编译：

1. 由于Term 继承自std::pair，Term 是一个新 类型。出于这个原因，您需要将BOOST_FUSION_ADAPT_STRUCT 应用到Term，无论是否已经在<boost/fusion/adapted/std_pair.hpp> 中为std::pair 完成了此操作：

   BOOST_FUSION_ADAPT_STRUCT(
       Term,
       (Factor, first)
       (std::vector<OpAndFactor>, second)
   )
   

   或者，您可以将Term 设为具有两个成员的独立结构，然后在其上应用BOOST_FUSION_ADAPT_STRUCT：

   struct Term { Factor f; std::vector<OpAndFactor>  o;};
   
   BOOST_FUSION_ADAPT_STRUCT(
       Term,
       (Factor, f)
       (std::vector<OpAndFactor>, o)
   )
   

   顺便说一句：你必须在这里完全限定std::vector，因为下面的will not compile：

   using std::vector;
   BOOST_FUSION_ADAPT_STRUCT(
       Term,
       (Factor, f)
       (vector<OpAndFactor>, o)
   )
   

2. 声明term_rule时使用Term而不是RWTerm：

   rule<It, Term(), space_type> term_rule;
   

完整代码：

#include <string>
#include <vector>
#include <utility>

#include <boost/spirit/include/qi.hpp>

#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/adapted.hpp>
#include <boost/fusion/include/adapted.hpp>

#include <boost/variant/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>

using boost::spirit::qi::grammar;
using boost::spirit::qi::space_type;
using boost::spirit::qi::rule;

struct Term; // forward dec
typedef boost::recursive_wrapper<Term> RWTerm;
typedef std::pair<char, RWTerm> OpAndRWTerm;
typedef std::pair<RWTerm, std::vector<OpAndRWTerm> > Expr;
typedef boost::variant<Expr, double> Factor;
typedef std::pair<char, Factor> OpAndFactor;
struct Term : public std::pair<Factor, std::vector<OpAndFactor> >{};

BOOST_FUSION_ADAPT_STRUCT(
    Term,
    (Factor, first)
    (std::vector<OpAndFactor>, second)
)


template<typename It>
struct formula_parser : grammar<It, Expr(), space_type> {
  formula_parser() : formula_parser::base_type(expr_rule) {
    using boost::spirit::qi::double_;
    using boost::spirit::ascii::char_;

    factor_rule %= double_ | parenthesis_rule;
    parenthesis_rule %= '(' >> expr_rule >> ')';
    op_and_factor_rule %= char_("/*") >> factor_rule;
    term_rule %= factor_rule >> *op_and_factor_rule;
    op_and_term_rule %= char_("+-") >> term_rule;
    expr_rule %= term_rule >> *op_and_term_rule;
  }
  rule<It, OpAndRWTerm(), space_type> op_and_term_rule;
  rule<It, Expr(), space_type> expr_rule;
  rule<It, OpAndFactor(), space_type> op_and_factor_rule;
  rule<It, Term(), space_type> term_rule;
  rule<It, Expr(), space_type> parenthesis_rule;
  rule<It, Factor(), space_type> factor_rule;
};

int main() {
  formula_parser<std::string::const_iterator> grammar;
}

live example