C++ 使用Boost.Spirit解析嵌套列表

我试图用Boost.Spirit解析嵌套的数字列表。这就是我到目前为止所做的：

//define a test string
std::string string = "[[\t1 , 2 ], [3, 4, 65.4]]";
auto it = string.begin();

//parse the string
std::vector<std::vector<double>> vector;
auto listRule = "[" >> (qi::double_ % ",") >> "]";
auto list2Rule = "[" >> (listRule % ",") >> "]";
bool match = qi::phrase_parse(it, string.end(), list2Rule, ascii::space, vector);

//check if we have a match
std::cout << "matched: " << std::boolalpha << match << '\n';
if (it != string.end())
    std::cout << "unmatched part: " << std::string{it, string.end()} << '\n';

//print the result
std::cout << "result\n";
for (const auto& v : vector) {
    std::cout << "[";
    for (double i : v)
        std::cout << i << ",";
    std::cout << "]\n";
}

我面临的问题是，它不接受空列表。例如，通过如下方式更改字符串：

std::string string = "[[\t1 , 2 ], [3, 4, 65.4], []]";

---

然后我没有匹配项（即

match==false

和

it==string.begin（）

）。显然，向量仍然会被填充，但最后一个空列表丢失了。有人能解释一下为什么会出现这种情况，以及如何解决它吗？

您正在使用Qi域中的原始表达式模板的

auto

——99.9%的情况下是未定义的行为：

现在，在修复该问题的同时，还要使列表正文成为可选的：

#include <boost/spirit/include/qi.hpp>
namespace qi = boost::spirit::qi;

int main() {
    using It      = std::string::const_iterator;
    using Skipper = qi::space_type;

    for(std::string const input : { "[[\t1 , 2 ], [3, 4, 65.4]]", "[[\t1 , 2 ], [3, 4, 65.4], []]", "[]" })
    {
        std::cout << " ---- '" << input << "' ----\n";
        auto it = input.begin();

        //parse the string
        using doubles = std::vector<double>;
        using vectors = std::vector<doubles>;

        qi::rule<It, doubles(), Skipper> doubles_ = "[" >> -(qi::double_ % ",") >> "]";
        qi::rule<It, vectors(), Skipper> vectors_ = "[" >> -(doubles_    % ",") >> "]";

        vectors data;
        bool match = qi::phrase_parse(it, input.end(), vectors_, qi::space, data);

        //check if we have a match
        std::cout << "matched: " << std::boolalpha << match << '\n';
        if (it != input.end())
            std::cout << "unmatched part: " << std::string{it, input.end()} << '\n';

        //print the result
        std::cout << "result\n";
        for (const auto& v : data) {
            std::cout << "[";
            for (double i : v)
                std::cout << i << ",";
            std::cout << "]\n";
        }
    }
}

---

%

解析器定义为：“List.Parse a delimited b一次或多次”。IIRC，您可以简单地将列表标记为可选，它将执行您想要的操作：

“[”>>-（listRule%，”>>“]嗯。。不，似乎不起作用。在这种情况下（同时使用
listRule
和
list2Rule
中的所有组合，我得到了一个向量列表，每个向量只有一个值…谢谢你的回答。但是，这不是我所期望的输出…请看我修改过的问题。该死。下次我应该更加注意输出。显然，这是不对的：）固定的。问题是在这种情况下，属性兼容性的魔力太大了：）是的，效果非常好！因此缺少三点：（a）将
auto
与规则一起使用是未定义的行为，（b）可以将解析器设置为可选的
-
，（c）根据解析器的实际工作方式，每个规则都可以有自己的属性类型。谢谢！
#include <boost/spirit/include/qi.hpp>
namespace qi = boost::spirit::qi;

int main() {
    using It      = std::string::const_iterator;
    using Skipper = qi::space_type;

    for(std::string const input : { "[[\t1 , 2 ], [3, 4, 65.4]]", "[[\t1 , 2 ], [3, 4, 65.4], []]", "[]" })
    {
        std::cout << " ---- '" << input << "' ----\n";
        auto it = input.begin();

        //parse the string
        using doubles = std::vector<double>;
        using vectors = std::vector<doubles>;

        qi::rule<It, doubles(), Skipper> doubles_ = "[" >> -(qi::double_ % ",") >> "]";
        qi::rule<It, vectors(), Skipper> vectors_ = "[" >> -(doubles_    % ",") >> "]";

        vectors data;
        bool match = qi::phrase_parse(it, input.end(), vectors_, qi::space, data);

        //check if we have a match
        std::cout << "matched: " << std::boolalpha << match << '\n';
        if (it != input.end())
            std::cout << "unmatched part: " << std::string{it, input.end()} << '\n';

        //print the result
        std::cout << "result\n";
        for (const auto& v : data) {
            std::cout << "[";
            for (double i : v)
                std::cout << i << ",";
            std::cout << "]\n";
        }
    }
}

 ---- '[[   1 , 2 ], [3, 4, 65.4]]' ----
matched: true
result
[1,2,]
[3,4,65.4,]
 ---- '[[   1 , 2 ], [3, 4, 65.4], []]' ----
matched: true
result
[1,2,]
[3,4,65.4,]
[]
 ---- '[]' ----
matched: true
result