Boost Spirit分析器无法编译

Question

我试图用boost :: spirit编写一个简单的表达式解析器。 我从计算器示例开始（请参阅： http : //www.boost.org/doc/libs/1_41_0/libs/spirit/example/qi/calc2_ast.cpp ），并尝试添加“ ref”规则来表示对一个变量。 这是我的代码：

#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/variant/recursive_variant.hpp>
#include <boost/variant/apply_visitor.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_function.hpp>

#include <iostream>
#include <vector>
#include <string>
namespace client{


    namespace qi = boost::spirit::qi;
    namespace ascii = boost::spirit::ascii;

    ///////////////////////////////////////////////////////////////////////////
    //  Our AST
    ///////////////////////////////////////////////////////////////////////////
    struct binary_op;
    struct unary_op;
    struct ref {
       std::string name;
       ref(const std::string& name) : name(name){}
    };
    struct nil {};

    struct expression_ast
    {
        typedef
            boost::variant<
                nil // can't happen!
              , unsigned int
              , ref
              , boost::recursive_wrapper<expression_ast>
              , boost::recursive_wrapper<binary_op>
              , boost::recursive_wrapper<unary_op>
            >
        type;

        expression_ast()
          : expr(nil()) {}

        template <typename Expr>
        expression_ast(Expr const& expr)
          : expr(expr) {}

        expression_ast& operator+=(expression_ast const& rhs);
        expression_ast& operator-=(expression_ast const& rhs);
        expression_ast& operator*=(expression_ast const& rhs);
        expression_ast& operator/=(expression_ast const& rhs);

        type expr;
    };



    struct binary_op
    {
        binary_op(
            char op
          , expression_ast const& left
          , expression_ast const& right)
        : op(op), left(left), right(right) {}

        char op;
        expression_ast left;
        expression_ast right;
    };

    struct unary_op
    {
        unary_op(
            char op
          , expression_ast const& subject)
        : op(op), subject(subject) {}

        char op;
        expression_ast subject;
    };

    expression_ast& expression_ast::operator+=(expression_ast const& rhs)
    {
        expr = binary_op('+', expr, rhs);
        return *this;
    }

    expression_ast& expression_ast::operator-=(expression_ast const& rhs)
    {
        expr = binary_op('-', expr, rhs);
        return *this;
    }

    expression_ast& expression_ast::operator*=(expression_ast const& rhs)
    {
        expr = binary_op('*', expr, rhs);
        return *this;
    }

    expression_ast& expression_ast::operator/=(expression_ast const& rhs)
    {
        expr = binary_op('/', expr, rhs);
        return *this;
    }

    // We should be using expression_ast::operator-. There's a bug
    // in phoenix type deduction mechanism that prevents us from
    // doing so. Phoenix will be switching to BOOST_TYPEOF. In the
    // meantime, we will use a phoenix::function below:
    struct negate_expr
    {
        template <typename T>
        struct result { typedef T type; };

        expression_ast operator()(expression_ast const& expr) const
        {
            return expression_ast(unary_op('-', expr));
        }
    };

   struct ref_maker
   {
      template <typename T>
      struct result { typedef T type; };

      ref operator()(std::string const& expr) const
      {
          return expr;
      }
   };


    boost::phoenix::function<negate_expr> neg;
    boost::phoenix::function<ref_maker> make_ref;

///////////////////////////////////////////////////////////////////////////
//  Walk the tree
///////////////////////////////////////////////////////////////////////////
struct ast_print
{
   typedef void result_type;

   void operator()(qi::info::nil) const {
   }
   void operator()(int n) const {
      std::cout << n;
   }

   void operator()(expression_ast const& ast) const {
      boost::apply_visitor(*this, ast.expr);
   }

   void operator()(binary_op const& expr) const {
      std::cout << "op:" << expr.op << "(";
      boost::apply_visitor(*this, expr.left.expr);
      std::cout << ", ";
      boost::apply_visitor(*this, expr.right.expr);
      std::cout << ')';
   }

   void operator()(unary_op const& expr) const {
      std::cout << "op:" << expr.op << "(";
      boost::apply_visitor(*this, expr.subject.expr);
      std::cout << ')';
   }

   void operator()(ref const& expr) const {
      std::cout << "ref:" << expr.name ;
   }
};

    ///////////////////////////////////////////////////////////////////////////
    //  Our calculator grammar
    ///////////////////////////////////////////////////////////////////////////
    template <typename Iterator>
    struct calculator : qi::grammar<Iterator, expression_ast(), ascii::space_type>
    {
        calculator() : calculator::base_type(expression)
        {
            using qi::_val;
            using qi::_1;
            using qi::uint_;
            using qi::lit;
            using ascii::alnum;

            identifier = (alnum [_val += _1]);

            expression =
                term                            [_val = _1]
                >> *(   ('+' >> term            [_val += _1])
                    |   ('-' >> term            [_val -= _1])
                    )
                ;

            term =
                factor                          [_val = _1]
                >> *(   ('*' >> factor          [_val *= _1])
                    |   ('/' >> factor          [_val /= _1])
                    )
                ;

            factor =
                uint_                           [_val = _1]
                | identifier                    [_val = make_ref(_1)]
                |   '(' >> expression           [_val = _1] >> ')'
                |   ('-' >> factor              [_val = neg(_1)])
                |   ('+' >> factor              [_val = _1])
                ;
        }
        qi::rule<Iterator, std::string(), ascii::space_type> identifier;
        qi::rule<Iterator, expression_ast(), ascii::space_type>  expression, term, factor;
    };
}

///////////////////////////////////////////////////////////////////////////////
//  Main program
///////////////////////////////////////////////////////////////////////////////
int
main()
{
    std::cout << "/////////////////////////////////////////////////////////\n\n";
    std::cout << "Expression parser...\n\n";
    std::cout << "/////////////////////////////////////////////////////////\n\n";
    std::cout << "Type an expression...or [q or Q] to quit\n\n";

    using client::ascii::space;
    using client::expression_ast;
    using client::ast_print;

    typedef std::string::const_iterator iterator_type;
    typedef client::calculator<iterator_type> calculator;

    calculator calc; // Our grammar

    std::string str;
    while (std::getline(std::cin, str))
    {
        if (str.empty() || str[0] == 'q' || str[0] == 'Q')
            break;

        std::string::const_iterator iter = str.begin();
        std::string::const_iterator end = str.end();
        expression_ast ast;
        ast_print printer;
        bool r = phrase_parse(iter, end, calc, space, ast);

        if (r && iter == end)
        {
            std::cout << "-------------------------\n";
            std::cout << "Parsing succeeded\n";
            printer(ast);
            std::cout << "\n-------------------------\n";
        }
        else
        {
            std::string rest(iter, end);
            std::cout << "-------------------------\n";
            std::cout << "Parsing failed\n";
            std::cout << "stopped at: \": " << rest << "\"\n";
            std::cout << "-------------------------\n";
        }
    }

    std::cout << "Bye... :-) \n\n";
    return 0;
}

我对示例所做的唯一更改如下：

1. 在变体中添加ref结构和条目。
2. 添加一个identifier规则，该规则将单个变量名读取为string
3. 创建函子和凤凰函数make_ref以从string进行ref
4. 向factor规则添加一个identifier选项，该factor规则调用make_ref将identifier string转换为ref 。

该代码不起作用。 始终存在以下类型错误：

/usr/include/boost/spirit/home/phoenix/core/detail/function_eval.hpp:135:69: error: could not convert ‘bench::expr::ref_maker::operator()(const string&) const((*(const string*)boost::phoenix::detail::help_rvalue_deduction<std::basic_string<char> >((* &(& _0)->boost::spirit::argument<N>::eval<boost::phoenix::basic_environment<boost::fusion::vector1<std::basic_string<char>&>, boost::spirit::context<boost::fusion::cons<bench::expr::expression_ast&, boost::fusion::nil>, boost::fusion::vector0<> >, bool, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_> >((* & env))))))’ from ‘bench::expr::ref’ to ‘std::basic_string<char>’

凤凰似乎希望make_ref返回一个string而不是ref ，但这不是我想要的。 我在这里做错了什么？

Answer 1

ref_maker的结果ref_maker是ref ，与输入参数不同。 做了

struct ref_maker
{
   template <typename T>
   struct result { typedef ref type; }; // <-- here.

   ref operator()(std::string const& expr) const
   {
       return expr;
   }
};

另外，将main中的bench::expr::替换为client:: ，否则会出现名称空间bench::expr不存在的错误。