逻辑非运算符在提升::精神::QI 中不起作用

如果在 qi：：grammar 中，我使用此基本规则

expression = (boost::spirit::ascii::string("aaa"));

它将解析"AAA"，仅此而已

当我使用这个（注意！）时，它根本不解析任何内容，而我希望它在除"aaa"之外的所有内容上都成功

expression = !(boost::spirit::ascii::string("aaa"));

我会错过一些包含吗？我正在使用提升 1.54.0。

编辑：

抱歉，这有点草稿，我为我的第一次试验修改了计算器示例......

#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <iostream>
#include <string>
#include <boost/spirit/include/qi_lit.hpp>
#include <boost/spirit/include/qi_not_predicate.hpp>
/*
 * 
    __grammar_calculator.cpp
HEADERS += 
    __grammar_calculator.h
 */
namespace client
{
    namespace qi = boost::spirit::qi;
    namespace ascii = boost::spirit::ascii;
    ///////////////////////////////////////////////////////////////////////////
    //  Our calculator grammar
    ///////////////////////////////////////////////////////////////////////////
    template <typename Iterator>
    struct calculator : qi::grammar<Iterator, int(), ascii::space_type>
    {
        calculator() : calculator::base_type(expression)
        {
            using qi::_val;
            using qi::_1;
            using qi::uint_;
            using boost::spirit::qi::lit;
            using boost::spirit::ascii::string;
            expression = !(boost::spirit::ascii::string("aaa"));
        }
        qi::rule<Iterator, int(), ascii::space_type> expression, term, factor;
    };
}
///////////////////////////////////////////////////////////////////////////////
//  Main program
///////////////////////////////////////////////////////////////////////////////
int
main()
{
    std::cout << "/////////////////////////////////////////////////////////nn";
    std::cout << "Expression parser...nn";
    std::cout << "/////////////////////////////////////////////////////////nn";
    std::cout << "Type an expression...or [q or Q] to quitnn";
    using boost::spirit::ascii::space;
    typedef std::string::const_iterator iterator_type;
    typedef client::calculator<iterator_type> calculator;
    calculator calc; // Our grammar
    std::string str;
    int result;
    while (std::getline(std::cin, str))
    {
        if (str.empty() || str[0] == 'q' || str[0] == 'Q')
            break;
        std::string::const_iterator iter = str.begin();
        std::string::const_iterator end = str.end();
        bool r = phrase_parse(iter, end, calc, space, result);
        if (r && iter == end)
        {
            std::cout << "-------------------------n";
            std::cout << "Parsing succeededn";
            //std::cout << "result = " << result << std::endl;
            std::cout << "-------------------------n";
        }
        else
        {
            std::string rest(iter, end);
            std::cout << "-------------------------n";
            std::cout << "Parsing failedn";
            //std::cout << "stopped at: ": " << rest << ""n";
            std::cout << "-------------------------n";
        }
    }
    std::cout << "Bye... :-) nn";
    return 0;
}

编辑2：

同样的一个更干净一点：

#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <iostream>
#include <string>
#include <boost/spirit/include/qi_not_predicate.hpp>
namespace client
{
    namespace qi = boost::spirit::qi;
    namespace ascii = boost::spirit::ascii;
    template <typename Iterator>
    struct test : qi::grammar<Iterator>
    {
        test() : test::base_type(expression)
        {
            using boost::spirit::ascii::string;
            expression = (boost::spirit::ascii::string("aaa"));
        }
        qi::rule<Iterator> expression;
    };
}
int main()
{
    std::cout << "/////////////////////////////////////////////////////////nn";
    std::cout << "Expression parser...nn";
    std::cout << "/////////////////////////////////////////////////////////nn";
    std::cout << "Type an expression...or [q or Q] to quitnn";
    using boost::spirit::ascii::space;
    typedef std::string::const_iterator iterator_type;
    typedef client::test<iterator_type> test;
    test tester; // Our grammar
    std::string str;
    while (std::getline(std::cin, str))
    {
        if (str.empty() || str[0] == 'q' || str[0] == 'Q')
            break;
        std::string::const_iterator iter = str.begin();
        std::string::const_iterator end = str.end();
        bool r = phrase_parse(iter, end, tester, space);
        if (r && iter == end)
        {
            std::cout << "-------------------------n";
            std::cout << "Parsing succeededn";
            std::cout << "-------------------------n";
        }
        else
        {
            std::string rest(iter, end);
            std::cout << "-------------------------n";
            std::cout << "Parsing failedn";
            std::cout << "-------------------------n";
        }
    }
    std::cout << "Bye... :-) nn";
    return 0;
}

答：

问题来自测试：

if (r && iter == end)

正如所指出的，运算符没有消耗任何东西，所以 iter！=end

在下文中，Sehe提供了一些替代方案。