BNF带有自定义修改使用 Spirit 的解析器

问题描述

使用 here 中的 bnf 解析器，我尝试添加一个字段以作为 List 的属性读取。所以我所做的就是改变：

using List = std::list<Term>;

到

struct List : public std::list<Term>{
int number;
}

例如：

<code>   ::=  <letter><digit> 34 | <letter><digit><code> 23

所以这些数字被读取为 List 的属性。问题是我无法读取数字作为 List 的属性。

解决方法

我更喜欢组合而不是继承（出于多种原因）。

所以

BOOST_FUSION_ADAPT_STRUCT(Ast::List,terms,number)

与

_list       = +_term;

然后修改规则来自

_list       = +_term >> qi::uint_;

成为

//#define BOOST_SPIRIT_DEBUG
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/adapted.hpp>
#include <fmt/ranges.h>
#include <fmt/ostream.h>
#include <iomanip>
namespace qi = boost::spirit::qi;

namespace Ast {
    struct Name : std::string {
        using std::string::string;
        using std::string::operator=;

        friend std::ostream& operator<<(std::ostream& os,Name const& n) {
            return os << '<' << n.c_str() << '>';
        }
    };

    using Term = boost::variant<Name,std::string>;

    struct List {
        std::list<Term> terms;
        int number;

        friend std::ostream& operator<<(std::ostream& os,List const& l) {
            for (auto& t : l.terms)
                os << t;
            return os << " " << l.number;
        }
    };

    using Expression = std::list<List>;

    struct Rule {
        Name name; // lhs
        Expression rhs;
    };

    using Syntax = std::list<Rule>;
}

BOOST_FUSION_ADAPT_STRUCT(Ast::List,number)
BOOST_FUSION_ADAPT_STRUCT(Ast::Rule,name,rhs)

namespace Parser {
    template <typename Iterator>
    struct BNF: qi::grammar<Iterator,Ast::Syntax()> {
        BNF(): BNF::base_type(start) {
            using namespace qi;
            start = skip(blank) [ _rule % +eol ];

            _rule       = _rule_name >> "::=" >> _expression;
            _expression = _list % '|';
            _list       = +_term >> qi::uint_;
            _term       = _literal | _rule_name ;
            _literal    = '"' >> *(_character - '"') >> '"'
                        | "'" >> *(_character - "'") >> "'";
            _character  = alnum | char_("\"'| !#$%&()*+,./:;>=<?@]\\^_`{}~[-");
            _rule_name  = '<' >> (alpha >> *(alnum | char_('-'))) >> '>';

            BOOST_SPIRIT_DEBUG_NODES(
                (_rule)(_expression)(_list)(_term)
                (_literal)(_character)
                (_rule_name))
        }

    private:
        qi::rule<Iterator,Ast::Syntax()>     start;
        qi::rule<Iterator,Ast::Rule(),qi::blank_type> _rule;
        qi::rule<Iterator,Ast::Expression(),qi::blank_type> _expression;
        qi::rule<Iterator,Ast::List(),qi::blank_type> _list;
        // lexemes
        qi::rule<Iterator,Ast::Term()>       _term;
        qi::rule<Iterator,Ast::Name()>       _rule_name;
        qi::rule<Iterator,std::string()>     _literal;
        qi::rule<Iterator,char()>            _character;
    };
}

int main() {
    Parser::BNF<std::string::const_iterator> const parser;

    std::string const input =
        R"(<code>   ::=  <letter><digit> 34 | <letter><digit><code> 23
<letter> ::= "a" 1 | "b" 2 | "c" 3 | "d" 4 | "e" 5 | "f" 6 | "g" 7 | "h" 8 | "i" 9
<digit>  ::= "9" 10 | "1" 11 | "2" 12 | "3" 13 | "4" 14
    )";

    auto it = input.begin(),itEnd = input.end();

    Ast::Syntax syntax;
    if (parse(it,itEnd,parser,syntax)) {
        for (auto& rule : syntax)
            fmt::print("{} ::= {}\n",rule.name,fmt::join(rule.rhs," | "));
    } else {
        std::cout << "Failed\n";
    }

    if (it != itEnd)
        std::cout << "Remaining: " << std::quoted(std::string(it,itEnd)) << "\n";
}

足以得到你所描述的：

Live On Compiler Explorer

code ::= <letter><digit> 34 | <letter><digit><code> 23
letter ::= a 1 | b 2 | c 3 | d 4 | e 5 | f 6 | g 7 | h 8 | i 9
digit ::= 9 10 | 1 11 | 2 12 | 3 13 | 4 14
Remaining: "
    "

印刷品

.match?