c – 在语义动作中添加到精神qi符号表

按照boost::spirit::qi::symbols documentation的开头段落,我假设从语义动作中添加符号到qi ::符号并不太难.不幸的是,它似乎并不像我想象的那么简单.

以下测试代码表明了问题：

#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <string>

namespace qi = boost::spirit::qi;

typedef qi::symbols<char, unsigned int> constants_dictionary;

template <typename Iter> struct parser : public qi::grammar<Iter, qi::space_type> {

    parser(constants_dictionary &dict) : parser::base_type(start) {

        start = qi::lit("@") >> ((+qi::char_) >> qi::uint_)[dict.add(qi::_1, qi::_2)];
    }

    qi::rule<Iter> start;
};

int main() {
    constants_dictionary dict;
    parser<std::string::const_iterator> prsr(dict);
    std::string test = "@foo 3";
    parse(test.begin(), test.end(), prsr, qi::space);
}

从VS2010中提供与qi :: _ 2相关的类型错误：

C:\Users\k\Coding\dashCompiler\spirit_test.cpp(12) : error C2664: 'const boost::
spirit::qi::symbols<Char,T>::adder &boost::spirit::qi::symbols<Char,T>::adder::o
perator ()<boost::spirit::_1_type>(const Str &,const T &) const' : cannot conver
t parameter 2 from 'const boost::spirit::_2_type' to 'const unsigned int &'
        with
        [
            Char=char,
            T=unsigned int,
            Str=boost::spirit::_1_type
        ]
        Reason: cannot convert from 'const boost::spirit::_2_type' to 'const uns
igned int'
        No user-defined-conversion operator available that can perform this conv
ersion, or the operator cannot be called
        C:\Users\k\Coding\dashCompiler\spirit_test.cpp(10) : while compiling cla
ss template member function 'parser<Iter>::parser(constants_dictionary &)'
        with
        [
            Iter=std::_String_const_iterator<char,std::char_traits<char>,std::al
locator<char>>
        ]
        C:\Users\k\Coding\dashCompiler\spirit_test.cpp(21) : see reference to cl
ass template instantiation 'parser<Iter>' being compiled
        with
        [
            Iter=std::_String_const_iterator<char,std::char_traits<char>,std::al
locator<char>>
        ]

(为令人讨厌的VS2010错误风格道歉)

我应该使用什么语法来添加(以及稍后删除)此表中的符号？

解决方法:

这个问题has been answered before.但是,你发布的代码存在很多问题,所以我会逐一修复它们,以免你不必要地盯着错误信息页面.

工作代码(加上输出验证)是here on liveworkspace.org.

笔记：

>语义动作必须是凤凰演员,即你需要

> boost :: bind,phoenix :: bind,std :: bind
> phoenix :: lambda<>或者phoenix :: function<>
>一个函数指针或多态可调用对象(根据文档)

我推荐phoenix :: bind(在这种特殊情况下),我在下面给出

>解析器的队长和启动规则之间存在不匹配
> qi :: char_吃掉所有字符.结合船长,结果
 在解析失败时,因为(显然)值中的数字也是
 吃了qi :: char_.基于qi :: lexeme [qi :: graph],我向您展示了许多解决方案之一
>使用qi :: lexeme’绕过’船长(即阻止qi :: graph切割)
跨越空白,因为船长,好吧,跳过它)
> qi :: parse不带船长;使用qi :: phrase_parse(
它似乎起作用的原因是任何尾随的“可变参数”参数都是
绑定到解析器的公开属性,在本例中是
未指定,因此qi :: unused_type).
>如果你想直接传递test.begin()和test.end()
qi :: phrase_parse,你需要明确你想要const迭代器.该
更典型的解决方案是引入明确类型的变量
(例如,第一个和最后一个)

#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <string>

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

typedef qi::symbols<char, unsigned int> constants_dictionary;

template <typename Iter> struct parser : qi::grammar<Iter, qi::space_type> 
{
    parser(constants_dictionary &dict) : parser::base_type(start) 
    {
        start = qi::lit("@") >> (qi::lexeme [+qi::graph] >> qi::uint_)
            [ phx::bind(dict.add, qi::_1, qi::_2) ]
            ;
    }

    qi::rule<Iter, qi::space_type> start;
};

int main() {
    constants_dictionary dict;
    parser<std::string::const_iterator> prsr(dict);
    const std::string test = "@foo 3";

    if (qi::phrase_parse(test.begin(), test.end(), prsr, qi::space))
    {
        std::cout << "check: " << dict.at("foo") << "\n";
    }
}

标签：boost-spirit-qi,c,boost-spirit
来源： https://codeday.me/bug/20190826/1726989.html