当以有效关键字(符号)开头时，如何防止 Boost Spirit 符号解析器接受关键字(符号)。我希望该构造无法将“ONEMORE”作为一个整体进行解析，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

这就是我喜欢的样子:

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，用于说明要点。

#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;
}

最佳答案

看qi::repository::distinct或者自己采取一些措施:

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;

规则声明为

qi::rule<Iterator, ascii::space_type> start;

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

查看 Live On Coliru

打印:

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;
}

关于c++ - 防止 Boost Spirit Symbol 解析器过早接受关键字，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21960167/