防止提升精神符号解析器过早接受关键字

Prevent the Boost Spirit Symbol parser from accepting a keyword too early

本文关键字：关键字 符号      更新时间：2023-10-16

当以有效关键字（符号）开头时，如何防止提升精神符号解析器接受关键字（符号）。我希望该结构无法将"ONEMORE"作为一个整体解析，并且无法成功解析"ONE"，因为这是一个有效的关键字，然后在"MORE"上失败。

下面是代码的实际输出：

Keyword as a number: 1
Keyword as a number: 2
Keyword as a number: 1
Invalid keyword: MORETHREE

这就是我喜欢的：

Keyword as a number: 1
Keyword as a number: 2
Invalid keyword: ONEMORE
Keyword as a number: 3

该代码只是一个示例来说明这一点。

#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <iostream>
#include <string>
using namespace std;
namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;
void printNumber( unsigned u )
{
    cout << "Keyword as a number: " << u << endl;
}
void printInvalidKeyword( const string &s )
{
    cout << "Invalid keyword: " << s << endl;
}
template <typename Iterator>
struct keyword_parser : qi::grammar<Iterator, ascii::space_type>
{
    struct mySymbols_ : qi::symbols<char, unsigned>
    {
        mySymbols_()
        {
            add
            ("ONE"   , 1)
            ("TWO"   , 2)
            ("THREE" , 2)
            ;
        }
    } mySymbols;
    keyword_parser() : keyword_parser::base_type(start)
    {
        using qi::_1;
        using qi::raw;
        using ascii::char_;
        start %= *(
                   mySymbols[&printNumber]
                    |
                   invalid[&printInvalidKeyword]
                   );
        invalid = +char_;
    }
    qi::rule<Iterator, ascii::space_type> start;
    qi::rule<Iterator, std::string(), ascii::space_type> invalid;
};
int main()
{
    using boost::spirit::ascii::space;
    typedef std::string::const_iterator iterator_type;
    typedef keyword_parser<iterator_type> keyword_parser;
    std::string s = "ONE TWO ONEMORE THREE";
    iterator_type b = s.begin();
    iterator_type e = s.end();
    phrase_parse(b, e, keyword_parser(), space);
    return 0;
}

看看qi::repository::distinct或自己采取一些措施：

start %= *(
           keyword  [cout << val("Keyword as a number: ") << _1 << endl]
         | invalid  [cout << val("Invalid keyword: ")     << _1 << endl]
         );
keyword = mySymbols >> !(char_("a-zA-Z0-9_"));
invalid = +ascii::graph;

声明为

qi::rule<Iterator, ascii::space_type> start;
// lexemes do not ignore embedded skippables
qi::rule<Iterator, int()> keyword;
qi::rule<Iterator, std::string()> invalid;

在科里鲁现场观看

指纹：

Keyword as a number: 1
Keyword as a number: 2
Invalid keyword: ONEMORE
Keyword as a number: 2

完整来源：

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <iostream>
#include <string>
using namespace std;
namespace qi    = boost::spirit::qi;
namespace phx   = boost::phoenix;
namespace ascii = boost::spirit::ascii;
template <typename Iterator>
struct keyword_parser : qi::grammar<Iterator, ascii::space_type>
{
    struct mySymbols_ : qi::symbols<char, unsigned>
    {
        mySymbols_()
        {
            add
            ("ONE"   , 1)
            ("TWO"   , 2)
            ("THREE" , 2)
            ;
        }
    } mySymbols;
    keyword_parser() : keyword_parser::base_type(start)
    {
        using qi::_1;
        using ascii::char_;
        using phx::val;
        start %= *(
                   keyword  [cout << val("Keyword as a number: ") << _1 << endl]
                 | invalid  [cout << val("Invalid keyword: ")     << _1 << endl]
                 );
        keyword = mySymbols >> !(char_("a-zA-Z0-9_"));
        invalid = +ascii::graph;
    }
    qi::rule<Iterator, ascii::space_type> start;
    // lexemes do not ignore embedded skippables
    qi::rule<Iterator, int()> keyword;
    qi::rule<Iterator, std::string()/*IMPLICIT LEXEME:, ascii::space_type*/> invalid;
};
int main()
{
    using boost::spirit::ascii::space;
    typedef std::string::const_iterator iterator_type;
    typedef keyword_parser<iterator_type> keyword_parser;
    std::string s = "ONE TWO ONEMORE THREE";
    iterator_type b = s.begin();
    iterator_type e = s.end();
    phrase_parse(b, e, keyword_parser(), space);
    return 0;
}
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