如何使用qi创建通用解析器?

我正在尝试使用qi创建通用解析器元素，因为我不幸(必须支持MSVC)无法使用X3. 这个想法是要有一个模板化的结构:

I am attempting to create generic parser-elements using qi as I unfortunately (MSVC must be supported) can not use X3. The idea is to have a templated struct:

template<class T> struct parse_type;

我可以这样使用:

template<class T> T from_string(std::string const& s)
{
    T res;
    parse_type<T> t;
    ...
    if (phrase_parse(...,parse_type<T>(),...,t))
}

或像这样专门化

template<class T,class Alloc> 
struct parse_type<std::vector<T,Alloc>>
{
    // Parse a vector using rule '[' >> parse_type<T> % ',' > ']';
}

主要目的是允许轻松解析例如std :: tuple，boost :: optional和boost :: variant(由于qi的贪婪特性，最后一个不能自动生成.)

The primary purpose is to allow for easy parsing of e.g. std::tuple, boost::optional and boost::variant (The last one can not be automatic due to the greedy nature of qi).

我希望您能收到有关如何处理此问题的反馈.当前，我的结构基于qi :: grammar，但是X3不支持语法，并且MSVC编译它时我想使用X3，而且对于必须提供船长我也有些不自在. 一种替代方法是在parse_type中具有一个静态函数，该函数返回适当的规则.我正在考虑这是否是一种更清洁的方法?

I would appreciate feedback as to how approach this. Currently I base my struct on qi::grammar, but grammar is not supported in X3 and I would like to use X3 when MSVC compiles this, and I am also a little bit uncomfortable with having to provide the skipper. An alternative would be to have a static function in parse_type that returns the appropriate rule. I am considering if this is a cleaner approach?

任何反馈将不胜感激.

Update2:用在运行时失败的可编译示例替换了代码片段.这是代码:

Update2: Replaced code-snippet with compilable example that fails at runtime. Here is the code:

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

//  Support to simplify
using iter = std::string::const_iterator;
void print(std::vector<int> const& v)
{
    std::cout << '[';
    for (auto i: v) std::cout  << i << ',';
    std::cout << "]";
}

namespace qi = boost::spirit::qi;

//  My rule factory - quite useless if you do not specialise
template<class T> struct ps_rule;

//  An example of using the factory
template<class T>
T from_string(std::string const& s)
{
    T result;
    iter first { std::begin(s) };
    auto rule = ps_rule<T>::get();
    phrase_parse(first,std::end(s),rule,qi::space,result);
    return result;
}

//  Specialising rule for int
template<>
struct ps_rule<int> 
{
    static qi::rule<iter,int()> get() { return qi::int_; } 
};

//  ... and for std::vector (where the elements must have rules)
template<class T,class Alloc>
struct ps_rule<std::vector<T,Alloc>>
{
    static qi::rule<iter,std::vector<T,Alloc>()> get()
    {
        qi::rule<iter,std::vector<T,Alloc>()> res;
        res.name("Vector");
        res =
                qi::lit('{')
            >>  ps_rule<T>::get() % ','
            >>  '}';
        return res;
    }
};

int main()
{
    //  This one works like a charm.
    std::cout << ((from_string<int>("100") == 100) ? "OK\n":"Failed\n");

    std::vector<int> v {1,2,3,4,5,6};

    //  This one fails
    std::cout << ((from_string<std::vector<int>>("{1,2,3,4,5,6}") == v) ? "OK\n":"Failed\n");
}

代码在boost/function_template.hpp第766行中失败:

The code fails in boost/function_template.hpp line 766:

result_type operator()(BOOST_FUNCTION_PARMS) const
{
  if (this->empty())
    boost::throw_exception(bad_function_call());

  return get_vtable()->invoker
           (this->functor BOOST_FUNCTION_COMMA BOOST_FUNCTION_ARGS);
}

此代码是boost :: function4中的成员函数 ，boost :: fusion :: vector0>& ，boost :: spirit :: unused_type const&> 问题是get_vtable返回一个无效的指针.

This code is a member function in boost::function4 ,boost::fusion::vector0 > & ,boost::spirit::unused_type const&> and the problem is that get_vtable returns an invalid pointer.