Boost.Spirit 解析可选前缀

我正在尝试解析一串以空格分隔、可选标记的关键字。例如

descr:expense type:receivable customer 27.3

其中冒号之前的表达式是标签，并且它是可选的（即假定默认标签）。

我无法让解析器完成我想要的事情。我对一个版本做了一些小的修改典型的例子其目的是解析键/值对（很像 HTTP 查询字符串）。

typedef std::pair<boost::optional<std::string>, std::string> pair_type;
typedef std::vector<pair_type> pairs_type;

template <typename Iterator>
struct field_value_sequence_default_field
  : qi::grammar<Iterator, pairs_type()>
{
    field_value_sequence_default_field()
      : field_value_sequence_default_field::base_type(query)
    {
        query =  pair >> *(qi::lit(' ') >> pair);
        pair  = -(field >> ':') >> value;
        field = +qi::char_("a-zA-Z0-9");
        value = +qi::char_("a-zA-Z0-9+-\\.");
    }

    qi::rule<Iterator, pairs_type()> query;
    qi::rule<Iterator, pair_type()> pair;
    qi::rule<Iterator, std::string()> field, value;
};

然而，当我解析它时，当标签被省略时，optional<string>不是空/假。相反，它有该值的副本。该对的第二部分也具有价值。

如果未标记的关键字不能是标记（语法规则，例如有小数点），那么事情就会像我期望的那样工作。

我究竟做错了什么？这是 PEG 的概念错误吗？

相反，它有该值的副本。该对的第二部分也具有价值。

这是容器属性和回溯的常见陷阱：使用qi::hold, e.g. 理解 Boost.spirit 的字符串解析器

    pair  = -qi::hold[field >> ':'] >> value;

Complete sample Live On Coliru

#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/adapted/std_pair.hpp>
#include <boost/optional/optional_io.hpp>
#include <iostream>

namespace qi    = boost::spirit::qi;

typedef std::pair<boost::optional<std::string>, std::string> pair_type;
typedef std::vector<pair_type> pairs_type;

template <typename Iterator>
struct Grammar : qi::grammar<Iterator, pairs_type()>
{
    Grammar() : Grammar::base_type(query) {
        query =  pair % ' ';
        pair  = -qi::hold[field >> ':'] >> value;
        field = +qi::char_("a-zA-Z0-9");
        value = +qi::char_("a-zA-Z0-9+-\\.");
    }
  private:
    qi::rule<Iterator, pairs_type()> query;
    qi::rule<Iterator, pair_type()> pair;
    qi::rule<Iterator, std::string()> field, value;
};

int main()
{
    using It = std::string::const_iterator;

    for (std::string const input : {
            "descr:expense type:receivable customer 27.3",
            "expense type:receivable customer 27.3",
            "descr:expense receivable customer 27.3",
            "expense receivable customer 27.3",
    }) {
        It f = input.begin(), l = input.end();

        std::cout << "==== '" << input << "' =============\n";
        pairs_type data;
        if (qi::parse(f, l, Grammar<It>(), data)) {
            std::cout << "Parsed: \n";
            for (auto& p : data) {
                std::cout << p.first << "\t->'" << p.second << "'\n";
            }
        } else {
            std::cout << "Parse failed\n";
        }

        if (f != l)
            std::cout << "Remaining unparsed: '" << std::string(f,l) << "'\n";
    }
}

Printing

==== 'descr:expense type:receivable customer 27.3' =============
Parsed: 
 descr  ->'expense'
 type   ->'receivable'
--  ->'customer'
--  ->'27.3'
==== 'expense type:receivable customer 27.3' =============
Parsed: 
--  ->'expense'
 type   ->'receivable'
--  ->'customer'
--  ->'27.3'
==== 'descr:expense receivable customer 27.3' =============
Parsed: 
 descr  ->'expense'
--  ->'receivable'
--  ->'customer'
--  ->'27.3'
==== 'expense receivable customer 27.3' =============
Parsed: 
--  ->'expense'
--  ->'receivable'
--  ->'customer'
--  ->'27.3'