Question

我需要解析一个键值对，其中键本身是示例中的固定字符串lke'cmd'。不幸的是qi :: lit没有综合属性，并且qi :: char_没有解析固定的字符串。以下代码无法编译。执行后我需要那个result.name == cmd。

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

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

struct CommandRuleType
{
  std::string name;
  int arg;
};

BOOST_FUSION_ADAPT_STRUCT(CommandRuleType, name, arg)

int main() {
    qi::rule<std::string::const_iterator, CommandRuleType(), qi::space_type> rule = qi::lit("cmd") >> "=" >> qi::int_;

    for (std::string const s : {"cmd = 1" }) {
        std::cout << std::quoted(s) << " -> ";
        CommandRuleType result;
        if (qi::phrase_parse(s.begin(), s.end(), rule, qi::space, result)) {
            std::cout << "result: " << result.name << "=" << result.arg << "\n";
        } else {
            std::cout << "parse failed\n";
        }
    }
}

Answer 1

qi::lit不公开属性。 qi::string做到了：

    rule = qi::string("cmd") >> "=" >> qi::int_;

Live On Coliru

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

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

struct CommandRuleType {
    std::string name;
    int arg;
};

BOOST_FUSION_ADAPT_STRUCT(CommandRuleType, name, arg)

int main() {
    qi::rule<std::string::const_iterator, CommandRuleType(), qi::space_type>
        rule = qi::string("cmd") >> "=" >> qi::int_;

    for (std::string const s : { "cmd = 1" }) {
        std::cout << std::quoted(s) << " -> ";
        CommandRuleType result;
        if (qi::phrase_parse(s.begin(), s.end(), rule, qi::space, result)) {
            std::cout << "result: " << result.name << "=" << result.arg << "\n";
        } else {
            std::cout << "parse failed\n";
        }
    }
}

打印

"cmd = 1" -> result: cmd=1

精神上如何解析字符串并将其用作返回值

1 个答案: