boost :: spirit :: qi :: parse - >沒有結果

Question

請考慮以下代碼：

namespace qi  = boost::spirit::qi;

typedef qi::rule<
    std::string::const_iterator
> rule_type;

rule_type value_rule = +qi::char_ - ( '[' | qi::eoi );

std::string input( "Hello World" );
std::string value0, value1;

bool b0 = qi::parse( input.begin( ),
                     input.end( ),
                     value_rule,
                     value0 );

bool b1 = qi::parse( input.begin( ),
                     input.end( ),
                     +qi::char_ - ( '[' | qi::eoi ),
                     value1 );

結果：

b0 = true  
b1 = true  
value0 = ""  
value1 = "Hello World"

我很困惑為什么結果不同。 為獲得相同的結果，qi :: rule類型的正確定義是什么？

Answer 1

您忘記使規則聲明其公開的屬性類型：

typedef qi::rule<std::string::const_iterator, std::string()> rule_type;

住在Coliru

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

namespace qi  = boost::spirit::qi;


int main() 
{
    std::string const input( "Hello World" );

    {
        typedef qi::rule<std::string::const_iterator, std::string()> rule_type;
        rule_type value_rule = +qi::char_ - ( '[' | qi::eoi );

        std::string value;
        bool ok = qi::parse( input.begin( ),
                input.end( ),
                value_rule,
                value );

        std::cout << std::boolalpha << ok << "\t" << value << "\n";
    }

    {
        std::string value;
        bool ok = qi::parse( input.begin( ),
                input.end( ),
                +qi::char_ - ( '[' | qi::eoi ),
                value );

        std::cout << std::boolalpha << ok << "\t" << value << "\n";
    }
}

產量

true    Hello World
true    Hello World