3

我有很多这样的规则:

cmd_BC = (dlm > timestamp > dlm > cid > dlm > double_)
         [
             _val = lazy_shared<dc::BoardControl>(_1, _2, _3)
         ];

我想让它更具可读性,例如:

cmd_BC = param(timestamp) > param(cid) > param(double_)

甚至

cmd_BC = params(timestamp, cid, double_)

正如sehe指出的那样,归结为有一些方法可以自动期望分隔符。这里有哪些选择?我自己,我看到了三种可能性,都有缺陷:

  1. 使用宏。这不允许使用较短的可变参数形式。
  2. 编写自定义前缀指令。我对Spirit的发条似乎没有足够的经验,但如果它真的不那么难,我会尝试。
  3. 编写一个包装函数。我试图不走运以下代码:

    template <typename T>
    auto param(const T & parser) -> decltype(qi::lit(dlm) > parser)
    {
        return qi::lit(dlm) > parser;
    }
    

    但它没有编译,失败了

        // report invalid argument not found (N is out of bounds)
        BOOST_SPIRIT_ASSERT_MSG(
            (N < sequence_size::value),
            index_is_out_of_bounds, ());
    

    我也尝试过return (...).alias(),但它也没有编译。

4

3 回答 3

2

这个解决方案不是很“精神”,不幸的是“需要 C++11”(我不确定如何在 c++03 中获得所需的结果类型),但它似乎有效。受此处示例的启发。

PS:哦,我没有看到你的编辑。你有几乎相同的例子。

更新:使用 _1、_2 和 _3 的语义操作添加了另一个测试
Update2:根据 vines 在评论中的建议更改了 operator() 和 operator[] 的签名。

更新 3:添加了构造组合解析器的可变参数 operator(),并在使用 boost::proto 找到更好的解决方案后删除了 operator[]。稍微改变了例子。

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/proto/proto.hpp>

namespace qi = boost::spirit::qi;
namespace proto = boost::proto;
namespace phx = boost::phoenix;
using namespace boost::proto;

//This is a proto grammar/transform that creates the prefixed parser. The parser created depends on the parser passed (if it's a kleene or not)
// in _make_xxx  "_" corresponds to the supplied parser and "_state" to the delimiter
struct CreatePrefixedParser: //you can use _make_greater instead of _make_shift_right if you want to use "expectation"
or_ <
    when < dereference<_>, //If it's a kleene parser...
        _make_shift_right ( //create the parser -> dlm >> *(parser -dlm)
            _state,
            _make_dereference (
                _make_minus ( _child_c<0> ( _ ),
                    _state ) ) ) > ,
    when < unary_plus<_>, //If it's a +parser
        _make_shift_right ( //create the parser -> dlm >> +(parser -dlm)
            _state,
            _make_unary_plus (
                _make_minus ( _child_c<0> ( _ ),
                    _state ) ) ) > ,
    otherwise < //if it's any other parser
        _make_shift_right ( //create the parser -> dlm >> (parser -dlm)
            _state,
            _make_minus ( _,
                _state ) ) >
> {};

//-------------------------------------------------------------
//this combines the parsers this way: parser1, parser2, parser3, parser4 -> parser1>>(parser2 >>(parser3 >> parser4)))
//you can use make_expr<tag::greater> if you want to use "expectation"
//I have absolutely no idea when "deep_copy" is required but it seems to work this way
template<typename Delim, typename First, typename ... Rest>
struct myparser
{
    static auto combine ( Delim dlm_, const First& first, const Rest&...rest ) ->
    decltype ( make_expr<tag::shift_right> ( CreatePrefixedParser() ( deep_copy ( first ), dlm_ ), myparser<Delim, Rest...>::combine ( dlm_, rest... ) ) )
    {
        return make_expr<tag::shift_right> ( CreatePrefixedParser() ( deep_copy ( first ), dlm_ ), myparser<Delim, Rest...>::combine ( dlm_, rest... ) );
    }

};

template<typename Delim, typename Last>
struct myparser<Delim, Last>
{

    static auto combine ( Delim dlm_, const Last& last ) -> decltype ( CreatePrefixedParser() ( deep_copy ( last ), dlm_ ) )
    {
        return CreatePrefixedParser() ( deep_copy ( last ), dlm_ );
    }
};
//-----------------------------------------------------------------

template <typename T>
struct prefixer
{
    T dlm_;
    prefixer ( T dlm ) : dlm_ ( dlm ) {}

    template <typename ... Args>
    auto operator() ( const Args&... args ) ->
    decltype ( deep_copy ( myparser<T, Args...>::combine ( dlm_, args... ) ) )
    {
        return deep_copy ( myparser<T, Args...>::combine ( dlm_, args... ) );
    }
};
template <typename T>
prefixer<T> make_prefixer ( T dlm )
{
    return prefixer<T> ( dlm );
}

int main()
{
    std::string test = "lameducklamedog";

    std::string::const_iterator f ( test.begin() ), l ( test.end() );

    auto param = make_prefixer ( qi::lit ( "lame" ) );
    qi::rule<std::string::const_iterator> dog = qi::lit ( "do" ) > qi::char_ ( 'g' );
    //qi::rule<std::string::const_iterator> duck = qi::lit ( "duck" ) | qi::int_;
    qi::rule<std::string::const_iterator,std::string()> quackdog = (param (*qi::alpha)  >> param( dog ));


     std::string what;
     if ( qi::parse ( f, l, quackdog, what ) && f == l )
         std::cout << "the duck and the dog are lame, specially the " << what  << std::endl;
     else
         std::cerr << "Uhoh\n" << std::string(f,l) << std::endl;

    test = "*-*2.34*-*10*-*0.16*-*12.5";
    std::string::const_iterator f2 ( test.begin() ), l2 ( test.end() );

    auto param2 = make_prefixer ( qi::lit ( "*-*" ) );
    double d;
    qi::rule<std::string::const_iterator> myrule = ( param2 ( qi::double_, qi::int_, qi::double_ , qi::double_) ) [phx::ref ( d ) = qi::_1 + qi::_2 + qi::_3 + qi::_4];

    if ( qi::parse ( f2, l2, myrule ) && f2 == l2 )
        std::cout << "the sum of the numbers is " << d << std::endl;
    else
        std::cerr << "Uhoh\n";

}
于 2012-10-01T18:49:51.610 回答
1

如果我理解正确,您正在寻找一种自动期望或忽略分隔符表达式 ( dlm) 的方法?

船长

这是精神上的船长的经典地形。如果分隔符是可变的,这尤其有用,例如空格(可以接受不同数量的空格);

bool ok = qi::phrase_parse(
      first, last,                // input iterators
      timestamp > cid > double_,  // just specify the expected params
      qi::space);                 // delimiter, e.g. any amount of whitespace

请注意使用phrase_parse启用带有跳过的语法。

%用解析器指令分隔

你可以明确地去界定语法:

 dlm     = qi::lit(','); // as an example, delimit by single comma
 rule    = timestamp > dlm > cid > dlm > double_;

这很乏味。可能对您有用的东西(取决于应执行的输入验证量:

 dlm     = qi::lit(','); // as an example, delimit by single comma
 rule    = (timestamp | cid | double_) % dlm;

(这将导致 的向量variant<timestampt_t, cid_t, double>

自己滚

您可以滚动自己的解析器指令,类似于karma::delimit,但用于输入。

Hartmut Kaiser 在这篇文档文章中概述了这个想法:

如果您有兴趣,我可以看看我是否可以将此作为示例(我以前没有使用过)。老实说,我很惊讶这样的东西还不存在,我认为它会成为Spirit Repository的主要候选者

于 2012-09-30T11:08:59.623 回答
1

这是我终于满意的解决方案。它基于以下三个答案:

我决定不让它接受任意二进制函子,因为我怀疑它在解析上下文中是否有任何实际用途。所以,

#include <boost/proto/deep_copy.hpp>

template <typename D>
struct prefixer
{
    template<typename... T>
    struct TypeOfPrefixedExpr;

    template<typename T>
    struct TypeOfPrefixedExpr<T>
    {
        typedef typename boost::proto::result_of::deep_copy
                 < decltype ( std::declval<D>() > std::declval<T>() ) >::type  type;
    };

    template<typename T, typename... P>
    struct TypeOfPrefixedExpr<T, P...>
    {
        typedef typename boost::proto::result_of::deep_copy
                 < decltype ( std::declval<D>() > std::declval<T>()
                              > std::declval<typename TypeOfPrefixedExpr<P...>::type>() ) >::type  type;
    };



    D dlm_;
    prefixer ( D && dlm ) : dlm_ ( dlm ) {}

    template <typename U>
    typename TypeOfPrefixedExpr<U>::type operator() (U && parser )
    {
        return boost::proto::deep_copy ( dlm_ > parser );
    }

    template <typename U, typename ... Tail>
    typename TypeOfPrefixedExpr<U, Tail...>::type
            operator() (U && parser, Tail && ... tail )
    {
        return boost::proto::deep_copy ( dlm_ > parser > (*this)(tail ...) );
    }
};

template <typename D>
prefixer<D> make_prefixer ( D && dlm )
{
    return prefixer<D> ( std::forward<D>(dlm) );
}

它是这样使用的:

auto params = make_prefixer(qi::lit(dlm));

cmd_ID      = params(string) [ _val = lazy_shared<dc::Auth>   (_1) ];

cmd_NAV     = params(timestamp, double_, double_, double_, double_, double_)
              [
                  _val = lazy_shared<dc::Navigation>( _1, _2, _3, _4, _5, _6 )
              ];

cmd_BC      = params(timestamp, cid, double_)
              [
                  _val = lazy_shared<dc::BoardControl>(_1, _2, _3)
              ];
于 2012-10-04T02:37:03.623 回答