Boost Spirit编译问题

Posted 2023-02-21

【中文标题】Boost Spirit编译问题

【英文标题】：Boost spirit compile issue

【发布时间】：2013-05-27 18:05:33

【问题描述】：

嗨，我是提升精神库的新手。你能告诉我为什么下面的代码不能编译吗？

当我在我的语法中添加“scientificNumber”规则时，它不会编译。可能是什么原因？ 我添加了“scientificNumber”规则，以便解析像“12E10”这样的科学记数法。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);
       

你能告诉我什么是 boost::spirit::qi::grammar

【问题讨论】：

一般来说，了解编译器输出对于查找错误很有帮助。所以也许你想告诉我们，编译器在抱怨什么（以及在哪里）？

因为它是一个模板错误，它太长而且实际上我不知道我应该在这里粘贴哪个部分。这就是为什么我没有把它放在这里。我认为我在定义 scienceNumber 规则并将其与语法结合时遇到了问题

您能否添加缺少的包含文件和“testExp”声明/定义，使其尽可能完整？

引人注目的是，这与这里的解析器基本相同：http://***.com/questions/15486531/boostspiritqi-defining-a-calculator-for-nullaries/15488691#15488691

【参考方案1】：

如果您只搜索包含 error 的第一行，您将看到以下注释：

            // 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：它准确地告诉你哪里出了问题。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lexical_cast inside Boost Spirit 来...解析一个数字（！？！）是一种讽刺。您可以测试原始的东西：它也可以与 LIT (qi::double_) 解析器一起工作。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';

【参考方案2】：

编译问题就在这里：

 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;

关于 boost::spirit::qi::grammar ，请查看：http://www.boost.org/doc/libs/1_53_0/libs/spirit/classic/doc/grammar.html

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