使用文字的默认值增强Spirit Qi整数 [英] Boost spirit qi integer with default value on literal

问题描述

我想用特定文字的默认值解析数字. 例如，如果程序接收到"nan"或"n/a"，则整数默认值为-1.

I want to parse numbers with default values on specific literals. For example if the program receives "nan" or "n/a" the integer default value will be -1.

// Example program
#include <iostream>
#include <string>

#include <boost/fusion/adapted.hpp>
#include <boost/fusion/include/adapted.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_fusion.hpp>
#include <boost/spirit/include/phoenix_object.hpp>

struct Person
{
    std::string name;
    int age;
    std::string gender;
};

BOOST_FUSION_ADAPT_STRUCT(
    Person,
    (std::string, name)(int, age)(std::string, gender)
)

template <typename Iterator>
    struct PersonGrammar : boost::spirit::qi::grammar<Iterator, Person(), boost::spirit::ascii::space_type>
{
    using Skipper = boost::spirit::ascii::space_type;

        PersonGrammar(std::ostream& error_stream = std::cerr)
            : PersonGrammar::base_type(start, "Person")
        {
            string = +boost::spirit::qi::char_("a-zA-Z0-9_ .()/-") | boost::spirit::qi::attr("(unspecified)");
            integer = boost::spirit::qi::lit("n/a") [boost::spirit::qi::_val = int(-1)] | boost::spirit::qi::int_;
            //integer = boost::spirit::qi::lexeme[boost::spirit::qi::no_case[boost::spirit::qi::eps > ( boost::spirit::qi::lit("n/a")[boost::spirit::qi::_val = -1] |  boost::spirit::qi::int_ ) ] ];

            const char sep = ',';

            start %= string >> boost::spirit::qi::lit(sep) >> integer >> boost::spirit::qi::lit(sep) >> string;

            string.name("string");
            integer.name("integer");

        }

private:
    boost::spirit::qi::rule<Iterator, std::string(), Skipper> string;
    boost::spirit::qi::rule<Iterator, int(), Skipper> integer;
    boost::spirit::qi::rule<Iterator, Person(), Skipper> start;
};

int main()
{
    std::string input_row = "Jon Snow, n/a, male";

    std::string::const_iterator iterator = input_row.begin();
    std::string::const_iterator end = input_row.end();

    PersonGrammar<std::string::const_iterator> grammar;
    Person person;

    bool success = boost::spirit::qi::phrase_parse(
        iterator, end, grammar, boost::spirit::ascii::space, person);

    std::cout << "Parse status: " << std::boolalpha << success << std::endl;

    std::cout << "Person: " << person.name << std::endl << "Age: " << person.age << std::endl << "Gender: " << person.gender << std::endl;

    return 0;
}

输出将是:

Parse status: true
Person: JonSnow
Age: -1
Gender: male

没关系.但我想用数字来填充它:

So it's okay. But I want to feed it with a number:

std::string input_row = "Jon Snow, 22, male";

输出将是:

Parse status: true
Person: JonSnow
Age: -2075995368
Gender: male

这是什么问题?我希望这不是一个错误.也许我使用的是错误的东西. 代码中有一个注释的词素.我也尝试过这种表达方式.不幸的是，它没有用.

What is the problem here? I hope this is not a bug. Maybe I'm using something wrong. There is a commented lexeme in the code. I tried that expression as well. Sadly it didn't work.

感谢您的帮助！

推荐答案

就是这样:

> (qi::int_ | (no_case["n/a"] >> qi::attr(-1)))

注意:

integer_ = lexeme[no_case[eps > ( lit("n/a")[_val = -1] |  int_ ) ] ];

缺少对int_的语义操作.或者，您需要使用%=强制进行自动传播.但是，我的第一个建议是更优雅一些，IMO.

lacks the semantic action on int_. OR you need to force automatic propagation with %=. But my first suggestion is way more elegant, IMO.

在魔盒上直播 >

#include <iostream>
#include <string>

#include <boost/fusion/adapted.hpp>
#include <boost/fusion/include/adapted.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_fusion.hpp>
#include <boost/spirit/include/phoenix_object.hpp>

struct Person
{
    std::string name;
    int age;
    std::string gender;
};

BOOST_FUSION_ADAPT_STRUCT(
    Person,
    (std::string, name)(int, age)(std::string, gender)
)

namespace qi = boost::spirit::qi;

template <typename Iterator>
    struct PersonGrammar : qi::grammar<Iterator, Person(), qi::ascii::space_type>
{
    PersonGrammar(std::ostream& /*error_stream*/ = std::cerr)
        : PersonGrammar::base_type(start, "Person")
    {
        using namespace qi;
        string_ = +char_("a-zA-Z0-9_ .()/-") | attr("(unspecified)");
        integer_ = int_ | no_case["n/a"] >> attr(-1);

        const char sep = ',';

        start = string_ >> sep >> integer_ >> sep >> string_;

        BOOST_SPIRIT_DEBUG_NODES((string_)(integer_))
    }

    private:
    using Skipper = boost::spirit::ascii::space_type;
    qi::rule<Iterator, std::string(), Skipper> string_;
    qi::rule<Iterator, int(), Skipper> integer_;
    qi::rule<Iterator, Person(), Skipper> start;
};

int main() {
    std::string input_row = "Jon Snow, n/a, male";

    std::string::const_iterator iterator = input_row.begin();
    std::string::const_iterator end = input_row.end();

    PersonGrammar<std::string::const_iterator> grammar;
    Person person;

    bool success = qi::phrase_parse(
        iterator, end, grammar, qi::ascii::space, person);

    std::cout << "Parse status: " << std::boolalpha << success << std::endl;
    std::cout << "Person: " << person.name << std::endl << "Age: " << person.age << std::endl << "Gender: " << person.gender << std::endl;
}

打印

Parse status: true
Person: JonSnow
Age: -1
Gender: male

这篇关于使用文字的默认值增强Spirit Qi整数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！