c – 用容器解析结构

如何使用boost.spirit x3解析为如下结构：

struct person{
    std::string name;
    std::vector<std::string> friends;
}

来自boost.spirit v2我会使用语法,但由于X3不支持语法,我不知道如何干净.

编辑：如果有人可以帮我编写一个解析字符串列表的解析器并返回第一个字符串的人是名称而字符串的res在朋友矢量中,那将是很好的.

解决方法:

使用x3解析比使用v2简单得多,因此移动时不会有太多麻烦.语法消失是件好事！

以下是如何解析字符串向量：

//#define BOOST_SPIRIT_X3_DEBUG

#include <fstream>
#include <iostream>
#include <string>
#include <type_traits>
#include <vector>

#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/include/io.hpp>
#include <boost/spirit/home/x3.hpp>
#include <boost/spirit/home/x3/support/ast/variant.hpp>

namespace x3 = boost::spirit::x3;

struct person
{
    std::string name;
    std::vector<std::string> friends;
};

BOOST_FUSION_ADAPT_STRUCT(
    person,
    (std::string, name)
    (std::vector<std::string>, friends)
);

auto const name = x3::rule<struct name_class, std::string> { "name" }
                = x3::raw[x3::lexeme[x3::alpha >> *x3::alnum]];

auto const root = x3::rule<struct person_class, person> { "person" }
                = name >> *name;

int main(int, char**)
{
    std::string const input = "bob john ellie";
    auto it = input.begin();
    auto end = input.end();

    person p;
    if (phrase_parse(it, end, root >> x3::eoi, x3::space, p))
    {
        std::cout << "parse succeeded" << std::endl;
        std::cout << p.name << " has " << p.friends.size() << " friends." << std::endl;
    }
    else
    {
        std::cout << "parse failed" << std::endl;
        if (it != end)
            std::cout << "remaining: " << std::string(it, end) << std::endl;
    }

    return 0;
}

如您所见,on Coliru,输出为：

parse succeeded
bob has 2 friends.

标签：boost-spirit-x3,c,c14,boost,boost-spirit
来源： https://codeday.me/bug/20190724/1525144.html