Boost spirit 鼓舞士气：有点像排列，但不完全是

我试图掌握精神，这意味着我是一个在这方面的傻瓜（因此，在下文中可能缺少合适的术语）

我必须分析这个：

value1 = 10
value2 = 20
value3 = 30
value4 = 40

顺序不重要，但每个“值1”。。。“value4”行必须正好出现一次。这样就可以了：

value1 = 10
value4 = 40
value2 = 20
value3 = 30

但这并不合适（重复的“值1”）：

也不包括以下内容（缺少“值4”）：

我怎样才能用精神做到这一点

额外问题：如果行“value3”是可选的，该怎么办

---

MaX.

我会使用置换解析器+附加的有效性检查

这样你就不必写所有的东西，而且仍然可以实现你想要的约束

#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/optional/optional_io.hpp>

struct X {
    boost::optional<int> value1;
    boost::optional<int> value2;
    boost::optional<int> value3;
    boost::optional<int> value4;
};

static bool is_valid(X const& x) { return x.value1 && x.value2 && x.value4; } // value3 is optional

BOOST_FUSION_ADAPT_STRUCT(X,
        (boost::optional<int>, value1)
        (boost::optional<int>, value2)
        (boost::optional<int>, value3)
        (boost::optional<int>, value4)
    )

int main() {
    namespace qi = boost::spirit::qi;
    namespace phx = boost::phoenix;
    using It = std::string::const_iterator;

    qi::rule<It, X(), qi::space_type> grammar;
    grammar =
        (("value1" > qi::lit('=') > qi::int_) ^
         ("value2" > qi::lit('=') > qi::int_) ^
         ("value3" > qi::lit('=') > qi::int_) ^
         ("value4" > qi::lit('=') > qi::int_))
        >> qi::eoi
        >> qi::eps(phx::bind(is_valid, qi::_val))
        ;

    for (std::string const& input : {
            "value1 = 10\nvalue2 = 20\nvalue3 = 30\nvalue4 = 40\n",
            // Order doesn't matter but each value1 ... value4 line must be present exactly once. This would be OK:
            "value1 = 10\nvalue4 = 40\nvalue2 = 20\nvalue3 = 30\n",
            // But this would not be OK (duplicated value1):
            "value1 = 10\nvalue2 = 20\nvalue3 = 30\nvalue4 = 40\nvalue1 = 10000\n",
            // Nor this (missing value4):
            "value1 = 10\nvalue2 = 20\nvalue3 = 30\n",
            // value3 _is_ optional though:
            "value1 = 10\nvalue2 = 20\nvalue4 = 40\n",
        })
    {
        std::cout << "---------------------------------------------------------\n";
        std::cout << "Parsing '" << input << "'\n";
        auto f(input.begin()), l(input.end());
        X parsed;
        bool ok = phrase_parse(f, l, grammar, qi::space, parsed);

        if (ok) {
            std::cout << "Parsing succeeded: " << boost::fusion::as_vector(parsed) << "\n";
        } else {
            std::cout << "Parsing failed\n";
        }

        if (f!=l)
            std::cout << "Remaing input '" << std::string(f,l) << "'\n";
    }
}

如您所见，最后一个问题解决了您的奖金问题

还可以查看Spirit存储库中的关键字解析器指令：

---

它允许您设置最小/最大出现次数

谢谢，关键字指令似乎符合要求。然而，在您的解决方案中，任何项目都可以是可选的，而我更感兴趣的是只选择其中一项。我说的对吗？@HometMaX你试过了吗？我想我已经说了为什么我会这样做：）无论如何，你可以自由使用关键字解析器（我没有太多的经验，我只知道它存在。它不是支持的Qi接口的一部分）嗨，sehe，不，我还没有试过。我希望在写下来之前100%地确定我理解了它（因为我必须解析的真实文件比上面的例子更复杂）。我对Fusion和Phoenix的使用也有点害怕（我一点也不知道）：我仍在思考是否值得学习它们，或者我是否可以不用它们就完成这项工作。但我不确定“不属于支持的Qi接口”意味着什么。谢谢你的帮助！干杯没关系。我想说的是，如果你要使用精神，那么学习融合和凤凰是值得的（反之亦然）。当然，您可以使用更重的枪（自定义指令和自定义点），但这要复杂得多。我会尽可能多地帮助你，所以我可以向你保证，没有他们，我可以在精神上完成大部分任务。

value1 = 10
value2 = 20
value3 = 30

#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/optional/optional_io.hpp>

struct X {
    boost::optional<int> value1;
    boost::optional<int> value2;
    boost::optional<int> value3;
    boost::optional<int> value4;
};

static bool is_valid(X const& x) { return x.value1 && x.value2 && x.value4; } // value3 is optional

BOOST_FUSION_ADAPT_STRUCT(X,
        (boost::optional<int>, value1)
        (boost::optional<int>, value2)
        (boost::optional<int>, value3)
        (boost::optional<int>, value4)
    )

int main() {
    namespace qi = boost::spirit::qi;
    namespace phx = boost::phoenix;
    using It = std::string::const_iterator;

    qi::rule<It, X(), qi::space_type> grammar;
    grammar =
        (("value1" > qi::lit('=') > qi::int_) ^
         ("value2" > qi::lit('=') > qi::int_) ^
         ("value3" > qi::lit('=') > qi::int_) ^
         ("value4" > qi::lit('=') > qi::int_))
        >> qi::eoi
        >> qi::eps(phx::bind(is_valid, qi::_val))
        ;

    for (std::string const& input : {
            "value1 = 10\nvalue2 = 20\nvalue3 = 30\nvalue4 = 40\n",
            // Order doesn't matter but each value1 ... value4 line must be present exactly once. This would be OK:
            "value1 = 10\nvalue4 = 40\nvalue2 = 20\nvalue3 = 30\n",
            // But this would not be OK (duplicated value1):
            "value1 = 10\nvalue2 = 20\nvalue3 = 30\nvalue4 = 40\nvalue1 = 10000\n",
            // Nor this (missing value4):
            "value1 = 10\nvalue2 = 20\nvalue3 = 30\n",
            // value3 _is_ optional though:
            "value1 = 10\nvalue2 = 20\nvalue4 = 40\n",
        })
    {
        std::cout << "---------------------------------------------------------\n";
        std::cout << "Parsing '" << input << "'\n";
        auto f(input.begin()), l(input.end());
        X parsed;
        bool ok = phrase_parse(f, l, grammar, qi::space, parsed);

        if (ok) {
            std::cout << "Parsing succeeded: " << boost::fusion::as_vector(parsed) << "\n";
        } else {
            std::cout << "Parsing failed\n";
        }

        if (f!=l)
            std::cout << "Remaing input '" << std::string(f,l) << "'\n";
    }
}

---------------------------------------------------------
Parsing 'value1 = 10
value2 = 20
value3 = 30
value4 = 40
'
Parsing succeeded: ( 10  20  30  40)
---------------------------------------------------------
Parsing 'value1 = 10
value4 = 40
value2 = 20
value3 = 30
'
Parsing succeeded: ( 10  20  30  40)
---------------------------------------------------------
Parsing 'value1 = 10
value2 = 20
value3 = 30
value4 = 40
value1 = 10000
'
Parsing failed
Remaing input 'value1 = 10
value2 = 20
value3 = 30
value4 = 40
value1 = 10000
'
---------------------------------------------------------
Parsing 'value1 = 10
value2 = 20
value3 = 30
'
Parsing failed
Remaing input 'value1 = 10
value2 = 20
value3 = 30
'
---------------------------------------------------------
Parsing 'value1 = 10
value2 = 20
value4 = 40
'
Parsing succeeded: ( 10  20 --  40)