Question

嗨，我很擅长提升精神图书馆。能不能让我知道为什么下面的代码不能编译？

当我将“scientificNumber”规则添加到我的语法中时，它不会编译。可能是什么原因？我添加了“scientificNumber”规则，以便能够解析像“12E10”这样的科学概念。我不知道这是否是正确的方法。

    namespace qi    = boost::spirit::qi;
    namespace phx   = boost::phoenix;

    typedef boost::function<double()> Value;

    #define BINARY_FUNCTOR(name, op)                        \
    struct name                                             \
    {                                                       \
        name(Value x, Value y): x_(x), y_(y) {}               \
        double operator()() { return x_() op y_(); }          \
        Value x_, y_;                                         \
    }; 

    BINARY_FUNCTOR(ADD, +)
    BINARY_FUNCTOR(SUB, -)
    BINARY_FUNCTOR(MUL, *)
    BINARY_FUNCTOR(DIV, /)

    struct LIT
    {
        LIT(double x): x_(x) {}
        double operator()() { return x_; }
        double x_;
    };

    struct NEG
    {
        NEG(Value x): x_(x) {}
        double operator()() { return -x_(); }
        Value x_;
    };

    struct SQRT
    {
        SQRT(Value x): x_(x){}
        double operator()() {return sqrt(x_()); }
        Value x_;
    };

    struct SCIENTIFIC
    {
        SCIENTIFIC(std::wstring x): x_(x){}
        double operator()() {return boost::lexical_cast<double>(x_); }
        std::wstring x_;
    };


    // expression grammar definition
    template <typename It, typename Skipper=boost::spirit::qi::space_type>
    struct parser : boost::spirit::qi::grammar<It, Value(),  Skipper>
    {
        parser() : parser::base_type(expression)
        {
            using namespace qi;
            expression =
                term                    [_val = _1]
                >> *( ('+' >> term  [_val = phx::construct<ADD>(_val, _1)])
                    | ('-' >> term  [_val = phx::construct<SUB>(_val, _1)])
                    );

            term =
                factor                [_val = _1]
                >> *( ('*' >> factor  [_val = phx::construct<MUL>(_val, _1)])
                    | ('/' >> factor  [_val = phx::construct<DIV>(_val, _1)])
                    );

            factor =
                double_               [_val = phx::construct<LIT>(_1)]
                |   scientificNumber  [_val = phx::construct<SCIENTIFIC>(_1)]
                |   '(' >> expression [_val = _1] >> ')'
                |   ('-' >> factor    [_val = phx::construct<NEG>(_1)])
                |   ('+' >> factor    [_val = _1])
                            |   (string("SQRT") >> '(' >> expression [_val = phx::construct<SQRT>(_1)] >> ')');

          scientificNumber = lexeme[+(boost::spirit::qi::digit) >> lit('E') >>  lit('-') >> +(boost::spirit::qi::digit)];


            BOOST_SPIRIT_DEBUG_NODE(expression);
            BOOST_SPIRIT_DEBUG_NODE(term);
            BOOST_SPIRIT_DEBUG_NODE(factor);
        }

      private:
        boost::spirit::qi::rule<It, std::wstring , Skipper> scientificNumber;
        qi::rule<It, Value(), Skipper> expression, term, factor;

    };



        int main()
        {

            std::wstring::const_iterator beginExpression(testExp.begin());
            std::wstring::const_iterator endExpression(testExp.end());
            typedef std::wstring::const_iterator It;
            parser<It , boost::spirit::qi::space_type> expressionParser;
            Value logicExpression;
   phrase_parse(beginExpression,endExpression,expressionParser,boost::spirit::qi::space,logicExpression);
       }

请你还能告诉我什么是boost :: spirit :: qi :: grammar

Answer 1

如果您只是搜索包含 error 的第一行，您会看到以下评论：

            // If you are seeing a compilation error here stating that the
            // fourth parameter can't be converted to a required target type
            // then you are probably trying to use a rule or a grammar with
            // an incompatible skipper type.
            if (f(first, last, context, skipper))

这是宾果游戏：它告诉你到底出了什么问题。它认为std::wstring是船长。

qi::rule<It, std::wstring, Skipper> scientificNumber; // huh?

根据文档，您需要将属性类型拼写为函数签名的返回类型：

qi::rule<It, std::wstring(), Skipper> scientificNumber;

现在，它编译并运行，见下文，打印输出：

Success: true   Value: 2.7e-09

重要

但是，使用 Boost Spirit中的lexical_cast 来解析数字（！？！） 是一个讽刺 。您可能只有测试原始内容：它也适用于LIT（qi::double_）解析器。请参阅http://ideone.com/mI1ESI

再次，请参阅此处的文档：link

完整代码

#include <boost/spirit/include/qi.hpp> #include <boost/spirit/include/phoenix.hpp> #include <boost/lexical_cast.hpp> namespace qi = boost::spirit::qi; namespace phx = boost::phoenix; typedef boost::function<double()> Value; #define BINARY_FUNCTOR(name, op) \ struct name \ { \ name(Value x, Value y): x_(x), y_(y) {} \ double operator()() { return x_() op y_(); } \ Value x_, y_; \ }; BINARY_FUNCTOR(ADD, +) BINARY_FUNCTOR(SUB, -) BINARY_FUNCTOR(MUL, *) BINARY_FUNCTOR(DIV, /) struct LIT { LIT(double x): x_(x) {} double operator()() { return x_; } double x_; }; struct NEG { NEG(Value x): x_(x) {} double operator()() { return -x_(); } Value x_; }; struct SQRT { SQRT(Value x): x_(x) {} double operator()() { return sqrt(x_()); } Value x_; }; struct SCIENTIFIC { SCIENTIFIC(std::wstring x): x_(x) {} double operator()() { return boost::lexical_cast<double>(x_); } std::wstring x_; }; // expression grammar definition template <typename It, typename Skipper=qi::space_type> struct parser : qi::grammar<It, Value(), Skipper> { parser() : parser::base_type(expression) { using namespace qi; expression = term [_val = _1] >> *(('+' >> term [_val = phx::construct<ADD>(_val, _1)]) | ('-' >> term [_val = phx::construct<SUB>(_val, _1)]) ); term = factor [_val = _1] >> *(('*' >> factor [_val = phx::construct<MUL>(_val, _1)]) | ('/' >> factor [_val = phx::construct<DIV>(_val, _1)]) ); factor = double_ [_val = phx::construct<LIT>(_1)] | scientificNumber [_val = phx::construct<SCIENTIFIC>(_1)] | '(' >> expression [_val = _1] >> ')' | ('-' >> factor [_val = phx::construct<NEG>(_1)]) | ('+' >> factor [_val = _1]) | (string("SQRT") >> '(' >> expression [_val = phx::construct<SQRT>(_1)] >> ')'); scientificNumber = lexeme[+(qi::digit) >> lit('E') >> lit('-') >> +(qi::digit)]; BOOST_SPIRIT_DEBUG_NODE(expression); BOOST_SPIRIT_DEBUG_NODE(term); BOOST_SPIRIT_DEBUG_NODE(factor); } private: qi::rule<It, std::wstring(), Skipper> scientificNumber; qi::rule<It, Value(), Skipper> expression, term, factor; }; int main() { const std::wstring testExp = L"3E-10*(12-3)"; typedef std::wstring::const_iterator It; It f(testExp.begin()), e(testExp.end()); parser<It, qi::space_type> expressionParser; Value logicExpression; bool ok = phrase_parse(f,e,expressionParser,qi::space,logicExpression); std::cout << "Success: " << std::boolalpha << ok << "\tValue: " << logicExpression() << '\n'; }

Answer 2

汇编问题就在这里：

 boost::spirit::qi::rule<It, std::wstring , Skipper> scientificNumber;

它不会产生任何属性，因为您不提供构造函数。这是修复：

 boost::spirit::qi::rule<It, std::wstring() , Skipper> scientificNumber;

对于boost :: spirit :: qi :: grammar，请检查：http://www.boost.org/doc/libs/1_53_0/libs/spirit/classic/doc/grammar.html

这里有很多例子：
https://stackoverflow.com/questions/tagged/boost-spirit

提升精神编译问题

2 个答案:

重要

完整代码