Optional is set even though optional_parser fails

[andreasbuhr]

In the following code, parsing a do-statement with a condition (heavily simplified), the std::optional representing the condition should not be set. However, it is. This is a bug in boost.parser.

void reproducer()
{
    namespace bp = boost::parser;

    constexpr auto condition_clause =
        bp::lit(U"while") > bp::lit(U"someexpression") >> bp::attr(true) ;

    constexpr auto do_statement =
        bp::lexeme[bp::lit(U"do") >> &bp::ws] > -condition_clause > bp::eol;


    {
        auto result = bp::parse(
            U"do\n"
            , do_statement, bp::blank, bp::trace::off);
        BOOST_TEST(result);
        std::optional<bool>& condition = result.value();
        BOOST_TEST(!condition.has_value());
    }
}

[andreasbuhr]

While the reproducer now works, I think the problem is not fundamentally solved.

The fundamental problem is that bp::lexeme[bp::lit(U"do") >> &bp::ws] tinkers with the std::optional it is given, even though it has no business doing so.
Your fix was to make sure the opt_parser writes to the attribute in any case. So it overwrites the wrong values written to the std::optional by bp::lexeme[bp::lit(U"do") >> &bp::ws]. But this fix only works if the bp::lexeme[bp::lit(U"do") >> &bp::ws] comes before the optional parser.

If the bp::lexeme[bp::lit(U"do") >> &bp::ws] comes after the optional parser, the problem is still there. The opt_parser defined by by -condition_clause writes correctly an empty std::optional. The bp::lexeme[bp::lit(U"do") >> &bp::ws] comes along an default-constructs a "bool" into the std::optional.

I extended the reproducer in #283