c++ - 增强::精神::保持空白

Question

我正在使用此代码将“k1 = v1; k2 = v2; k3 = v3; kn = vn”字符串解析为映射。

    qi::phrase_parse(
      begin,end,
      *(*~qi::char_('=') >> '=' >> *~qi::char_(';') >> -qi::lit(';')),
      qi::ascii::space, dict);

上面的代码将删除空格字符，例如“some_key = 1 2 3”变成some_key-> 123

我不知道如何删除或替换第四个参数:qi::ascii::space

基本上，我想在用'='分割后保留原始字符串(键和值)。

我对精神没有太多的经验/知识。确实需要投入时间来学习。

Answer 1

如果您不希望有船长，只需使用qi::parse而不是qi::phrase_parse即可:

qi::parse(
  begin,end,
  *(*~qi::char_(";=") >> '=' >> *~qi::char_(';') >> -qi::lit(';')),
  dict);

但是，您可能确实想有选择地跳过空格。最简单的方法通常是拥有一个通用的船长，然后标记lexeme区域(您不允许该船长的区域):

qi::phrase_parse(
  begin, end,
  *(qi::lexeme[+(qi::graph - '=')]
     >> '='
     >> qi::lexeme[*~qi::char_(';')] >> (qi::eoi|';')),
  qi::ascii::space, dict);

linked answer确实提供了更多有关如何在Qi中使用船长的技术/背景

演示时间

Live On Coliru

#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/adapted/std_pair.hpp>
#include <map>
#include <iomanip>
namespace qi = boost::spirit::qi;

int main() {
    for (std::string const& input : { 
            R"()",
            R"(foo=bar)",
            R"(foo=bar;)",
            R"( foo = bar ; )",
            R"( foo = bar ;
foo 
= qux; baz =

    quux 
corge grault
 thud

; x=)",
            // failing:
            R"(;foo = bar;)",
        })
    {
        std::cout << "-------------------------\n";
        auto f=begin(input), l=end(input);

        std::multimap<std::string, std::string> dict;

        bool ok = qi::phrase_parse(f, l,
          (qi::lexeme[+(qi::graph - '=' - ';')]
             >> '='
             >> qi::lexeme[*~qi::char_(';')]
          ) % ';',
          qi::space,
          dict);

        if (ok) {
            std::cout << "Parsed " << dict.size() << " elements:\n";
            for (auto& [k,v]: dict) {
                std::cout << " - " << std::quoted(k) << " -> " << std::quoted(v) << "\n";
            }
        } else {
            std::cout << "Parse failed\n";
        }

        if (f!=l) {
            std::cout << "Remaining input: " << std::quoted(std::string(f,l)) << "\n";
        }
    }

}

版画

-------------------------
Parse failed
-------------------------
Parsed 1 elements:
 - "foo" -> "bar"
-------------------------
Parsed 1 elements:
 - "foo" -> "bar"
Remaining input: ";"
-------------------------
Parsed 1 elements:
 - "foo" -> "bar "
Remaining input: "; "
-------------------------
Parsed 4 elements:
 - "baz" -> "quux 
corge grault
 thud

"
 - "foo" -> "bar "
 - "foo" -> "qux"
 - "x" -> ""
-------------------------
Parse failed
Remaining input: ";foo = bar;"