Boost.Spirit X3 解析器 "no type named type in(...)"

Question

我在玩 Boost.Spirit X3 计算器示例时遇到了一个我无法理解的错误。
我最小化了程序以降低复杂性仍然抛出相同的错误。假设我想将输入解析为语句列表（字符串），后跟定界符（';'）。

这是我的结构：

namespace client { namespace ast
 {    
     struct program
    {
        std::list<std::string> stmts;
    };
 }}

BOOST_FUSION_ADAPT_STRUCT(client::ast::program,
            (std::list<std::string>, stmts)
)

语法如下：

namespace client
{ 
    namespace grammar
    {

   x3::rule<class program, ast::program> const program("program");

    auto const program_def =
            *((*char_) > ';')
            ;

   BOOST_SPIRIT_DEFINE(
       program
    );
    auto calculator = program;
}

using grammar::calculator;

}

已调用

    int
    main()
    {
    std::cout <<"///////////////////////////////////////////\n\n";
    std::cout << "Expression parser...\n\n";
    std::cout << //////////////////////////////////////////////////\n\n";
    std::cout << "Type an expression...or [q or Q] to quit\n\n";

    typedef std::string::const_iterator iterator_type;
    typedef client::ast::program ast_program;

    std::string str;
    while (std::getline(std::cin, str))
    {
        if (str.empty() || str[0] == 'q' || str[0] == 'Q')
            break;

        auto& calc = client::calculator;    // Our grammar
        ast_program program;                // Our program (AST)

        iterator_type iter = str.begin();
        iterator_type end = str.end();
        boost::spirit::x3::ascii::space_type space;
        bool r = phrase_parse(iter, end, calc, space, program);

        if (r && iter == end)
        {
            std::cout << "-------------------------\n";
            std::cout << "Parsing succeeded\n";
            std::cout<< '\n';
            std::cout << "-------------------------\n";
        }
        else
        {
            std::cout << "-------------------------\n";
            std::cout << "Parsing failed\n";
            std::cout << "-------------------------\n";
        }
    }

    std::cout << "Bye... :-) \n\n";
    return 0;
}

我得到的错误是

/opt/boost_1_66_0/boost/spirit/home/x3/support/traits/container_traits.hpp: In instantiation of ‘struct boost::spirit::x3::traits::container_value<client::ast::program, void>’:
.
.
.

/opt/boost_1_66_0/boost/spirit/home/x3/support/traits/container_traits.hpp:76:12: error: no type named ‘value_type’ in ‘struct client::ast::program’
         struct container_value
/opt/boost_1_66_0/boost/spirit/home/x3/operator/detail/sequence.hpp:497:72: error: no type named ‘type’ in ‘struct boost::spirit::x3::traits::container_value<client::ast::program, void>’
          , typename traits::is_substitute<attribute_type, value_type>::type());
                                                                        ^~~~~~

我尝试过的事情：

正在关注 Getting boost::spirit::qi to use stl containers
尽管它使用 Qi 我仍然尝试过：

namespace boost{namespace spirit{ namespace traits{
template<>
struct container_value<client::ast::program> 
//also with struct container<client::ast::program, void>
{
      typedef std::list<std::string> type;
};
}}}

你看我有点蒙在鼓里，所以预计无济于事。

parser2.cpp:41:8: error: ‘container_value’ is not a class template
 struct container_value<client::ast::program>
        ^~~~~~~~~~~~~~~

在同一个问题中我作者说
"There is one known limitation though, when you try to use a struct that has a single element that is also a container compilation fails unless you add qi::eps >> ... to your rule."

我确实尝试添加一个虚拟 eps，但也没有成功。

请帮我解读一下错误的含义。

Answer 1

是的。这看起来像是涉及单元素序列时属性自动传播的另一个限制。

我可能会硬着头皮将规则定义从原来的（以及您期望的工作方式）更改为：

x3::rule<class program_, std::vector<std::string> >

这消除了混乱的根源。

其他说明：

你有 char_ 它也吃 ';' 所以语法 永远不会 成功因为没有 ';' 会跟随"statement".
你的语句不是词素，所以空格被丢弃（这是你的意思吗？见Boost spirit skipper issues）
你的语句可能是空的，这意味着解析总是会在输入结束时失败（它会总是读取一个空状态，然后发现预期的 ';' 丢失了）。通过在接受声明之前要求至少 1 个字符来修复它。

有一些 simplifications/style 更改：

Live On Coliru

#include <boost/fusion/adapted.hpp>
#include <boost/spirit/home/x3.hpp>
#include <list>

namespace x3 = boost::spirit::x3;

namespace ast {
    using statement = std::string;

    struct program {
        std::list<statement> stmts;
    };
} 

BOOST_FUSION_ADAPT_STRUCT(ast::program, stmts)

namespace grammar {
    auto statement 
        = x3::rule<class statement_, ast::statement> {"statement"}
        = +~x3::char_(';');
    auto program 
        = x3::rule<class program_, std::list<ast::statement> > {"program"}
        = *(statement >> ';');
} 

#include <iostream>
#include <iomanip>

int main() {
    std::cout << "Type an expression...or [q or Q] to quit\n\n";

    using It = std::string::const_iterator;

    for (std::string str; std::getline(std::cin, str);) {
        if (str.empty() || str[0] == 'q' || str[0] == 'Q')
            break;

        auto &parser = grammar::program;
        ast::program program; // Our program (AST)

        It iter = str.begin(), end = str.end();
        if (phrase_parse(iter, end, parser, x3::space, program)) {
            std::cout << "Parsing succeeded\n";
            for (auto& s : program.stmts) {
                std::cout << "Statement: " << std::quoted(s, '\'') << "\n";
            }
        }
        else
            std::cout << "Parsing failed\n";

        if (iter != end)
            std::cout << "Remaining unparsed: " << std::quoted(std::string(iter, end), '\'') << "\n";
    }
}

输入 "a;b;c;d;" 打印：

Parsing succeeded
Statement: 'a'
Statement: 'b'
Statement: 'c'
Statement: 'd'

Boost.Spirit X3 解析器 "no type named type in(...)"

Boost.Spirit X3 parser "no type named type in(...)"

c++

parsing

boost-spirit

这是我的结构：

语法如下：

已调用

我得到的错误是

我尝试过的事情：