c++ - boost spirit 和凤凰解析为 std::string

Question

如何解析成带有 boost spirit 和 phoenix 的 std::string？

例如，下面我成功地将 1234 解析为一个 int，但是当我尝试解析为一个字符串时，解析失败了。如果我将 qi::alnum 替换为 +qi::alnum 则解析成功但字符串编号具有不需要的值“4”。好像 qi::alnum 只匹配一个字符；如何解析多字符字母数字字符串？

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>

#include <iostream>
#include <string>

int main() {

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

    std::string s("--1234--");

    {
        int number = 0;
        bool success = qi::parse(s.begin(), s.end(),
              qi::lit("--") >> qi::int_[ phoenix::ref(number) = qi::_1 ] >> qi::lit("--")
            );
        std::cout << success << " qi::int_ parses: " << number << std::endl;
    }

    {
        std::string number;
        bool success = qi::parse(s.begin(), s.end(),
              qi::lit("--") >> qi::alnum[ phoenix::ref(number) = qi::_1 ] >> qi::lit("--")
            );
        std::cout << success << " qi::alpha parses: " << number << std::endl;
    }

    return 0;
}

Answer 1

使用qi::as_string ，例如

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>

#include <iostream>
#include <string>

int main() {

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

    std::string s("--1234--");

    std::string number;
    bool success = qi::parse( s.begin(), s.end(),
        qi::lit("--")
        >> qi::as_string [ +qi::alnum ] [ phoenix::ref(number) = qi::_1 ]
        >> qi::lit("--"));
    std::cout << success << " qi::as_string parses: " << number << std::endl;

    return 0;
}