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Why doesn't boost :: spirit :: qi :: parse () set boost :: variant?

When trying to parse text in boost :: variant, the value of the variant does not change. Parsers themselves work fine, so I guess I'm doing something wrong with the option code.

I am using boost 1.46.1 and the following code compiles in Visual Studio 2008.

First update

hkaiser noted that the arguments of the rule pattern and grammar should not be Variant, but Variant().
This has changed a bit since I now have a compilation error in boost_1_46_1\boost\variant\variant.hpp(1304). The comment says:

// NOTE TO USER :
// Compile error here indicates that the given type is not 
// unambiguously convertible to one of the variant types
// (or that no conversion exists).

Thus, it is obvious that the attribute of the expression is (qi::double_ | +qi::char_)not boost::variant<double, std::string>. But then what?

Second update

typedef boost::variant<double, std::vector<char>> Variant; . std::string...

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

int main()
{
    namespace qi = boost::spirit::qi;

    typedef std::string::const_iterator Iterator;

    const std::string a("foo"), b("0.5");

    // This works
    {
        std::string stringResult;
        Iterator itA = a.begin();
        const bool isStringParsed =
            qi::parse(itA, a.end(), +qi::char_, stringResult);

        double doubleResult = -1;
        Iterator itB = b.begin();
        const bool isDoubleParsed =
            qi::parse(itB, b.end(), qi::double_, doubleResult);

        std::cout
                << "A Parsed? " << isStringParsed <<
                ", Value? " << stringResult << "\n"
                << "B Parsed? " << isDoubleParsed <<
                ", Value? " << doubleResult << std::endl;

        // Output:
        // A Parsed? 1, Value? foo
        // B Parsed? 1, Value? 0.5
    }


    // This also works now
    {
        typedef boost::variant<double, std::vector<char>> Variant; // vector<char>, not string!

        struct variant_grammar : qi::grammar<Iterator, Variant()> // "Variant()", not "Variant"!
        {
            qi::rule<Iterator, Variant()> m_rule; // "Variant()", not "Variant"!

            variant_grammar() : variant_grammar::base_type(m_rule)
            {
                m_rule %= (qi::double_ | +qi::char_);
            }
        };

        variant_grammar varGrammar;

        Variant varA(-1), varB(-1);
        Iterator itA = a.begin();
        const bool isVarAParsed = qi::parse(itA, a.end(), varGrammar, varA);
        Iterator itB = b.begin();
        const bool isVarBParsed = qi::parse(itB, b.end(), varGrammar, varB);

        // std::vector<char> cannot be put into std::cout but
        // needs to be converted to a std::string (or char*) first.
        // The conversion I came up with is very ugly but it not the point
        // of this question anyway, so I omitted it.
        // You'll have to believe me here, when I'm saying it works..

        // Output:
        // A (variant): Parsed? 1, Value? foo, Remaining text = ''
        // B (variant): Parsed? 1, Value? 0.5, Remaining text = ''
    }

    return 0;
}
+5
1

:

qi::rule<Iterator, Variant()> m_rule;

, , ( , ).

+4

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