c++ - 使用 boost::spirit::qi 解析 double 列表

Question

我越来越熟悉 boost::spirit 并想像下面这样解析字符串:

double_1 | double_2 | ... | double_n | double_1% | double_2% ... | double_m%

其中 m>=0，n>=0。

例如，下面的所有行都应该解析正常:

91.3 | 44 | 5e-3 | 12% | 11%

91.3 | 44 | 5e-3

12% | 11%

我想使用boost::spirit::qi。

所以，我编写了如下两个解析器:

namespace client
{
    namespace qi = boost::spirit::qi;
    namespace ascii = boost::spirit::ascii;

    template <typename Iterator>
    bool parse_numbers(Iterator& first, Iterator last, std::vector<double>& v)
    {
        using qi::double_;
        using qi::phrase_parse;        
        using ascii::space;

        bool r = phrase_parse(first, last,

            //  Begin grammar
            (
                // double_ % '|'
                double_ >> *('|' >> double_ >> '|')
            )
            ,
            //  End grammar

            space, v);
        return r;
    }
    
    template <typename Iterator>
    bool parse_numbersWithPercents(Iterator& first, Iterator last, std::vector<double>& v)
    {
        using qi::double_;
        using qi::phrase_parse;
        using ascii::space;

        bool r = phrase_parse(first, last,

            //  Begin grammar
            (
                (double_ >> '%') % '|'
            )
            ,
            //  End grammar

            space, v);

        if (first != last) // fail if we did not get a full match
            return false;
        return r;
    }
}

然后，我像下面这样在 main 中调用它们:

int main()
{
    std::cout << "Give me a list of numbers in a format  double_1 | double_2 | ... | double_n | double_1% | double_2% ... | double_m%\n";
    std::cout << "The numbers will be inserted in a vector of numbers\n";
    std::cout << "Type [q or Q] to quit\n\n";

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

        std::vector<double> v;
        std::string::iterator begin = str.begin(), end = str.end();
        if (client::parse_numbers(begin, end, v))
        {
            std::cout << "-------------------------\n";
            std::cout << "First Part Parsing succeeded\n";

            for (std::vector<double>::size_type i = 0; i < v.size(); ++i)
                std::cout << i << ": " << v[i] << std::endl;

            std::cout << "\n-------------------------\n";
            if(begin != end) {
                if('|' == *begin) ++begin;
                if(begin != end) {
                    std::cout << "Parsing second part: " << std::string(begin, end) << std::endl;
                    std::vector<double> v1;
                     if (client::parse_numbersWithPercents(begin, end, v1))
                    {
                        std::cout << "-------------------------\n";
                        std::cout << "Second Part Parsing succeeded\n";

                        for (std::vector<double>::size_type i = 0; i < v1.size(); ++i)
                            std::cout << i << ": " << v1[i] << std::endl;

                        std::cout << "\n-------------------------\n";
                } else {
                    std::cout << "-------------------------\n";
                    std::cout << "Second Part Parsing failed\n";
                    std::cout << "-------------------------\n";
                    
                    if(begin != end) {
                        std::cout << "Remaining part is: " << std::string(begin, end) << std::endl; }
                    }
            }
        }
        }
        else
        {
            std::cout << "-------------------------\n";
            std::cout << "First Part Parsing failed\n";
            std::cout << "-------------------------\n";
            
            if(begin != end) {
                std::cout << "Remaining part is: " << std::string(begin, end) << std::endl; }
        }
    }

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

如您所见，此方法不适用于极端情况，例如:

91.3 | 44 | 5e-3

12% | 11%

我很感兴趣是否有另一种方法可以使用 boost 库以更简单的方式执行相同的操作。或者以某种方式更正我的解析器以正确解析上述极端情况。最好将第一部分和第二部分放在单独的容器中。

提前致谢。

Answer 1

哈哈。我的直觉是这应该非常简单。然而，我得出的结论是，这确实有点不平凡。

问题在于将非列表重复分隔符设为可选。我苦思冥想了很久，想出了让它成为可选的最优雅的方式，并想出了这个:

Live On Coliru

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

namespace qi = boost::spirit::qi;

namespace {
    using double_vec = std::vector<double>;
    using It         = std::string::const_iterator;

    static const qi::rule<It, double_vec(bool percent), qi::blank_type> doubles_
        = (qi::double_ >> (qi::eps(qi::_r1) >> '%' | !qi::lit('%'))) % '|';
}

int main() {
    std::string str;
    while (std::getline(std::cin, str)) {
        It f = str.begin(), l = str.end();

        double_vec v, w;

        bool ok = qi::phrase_parse(f, l, 
                  (doubles_(false) >> -('|' >> doubles_(true))) 
                | qi::attr(double_vec{}) >> doubles_(true),
                qi::blank, v, w);

        if (ok && f == l) {
            std::cout << "Parsed " << v.size() << "/" << w.size() << " elements\n";
        } else {
            std::istringstream iss(str);
            if (iss >> str && (str == "q" || str == "Q"))
                break;
            std::cout << "Invalid input. Remaining '" << std::string(f,l) << "'\n";
        }
    }
}

在给定测试输入的情况下会产生以下结果:

./test <<INPUT
91.3 | 44 | 5e-3 | 12% | 11% 
91.3 | 44 | 5e-3 
12% | 11%
q
INPUT
Parsed 3/2 elements
Parsed 3/0 elements
Parsed 0/2 elements

根据你在这里试图/实际/实现的目标，事情可能会更优雅

更新 作为对评论的回应，以下是我通过放宽语法来实际改进的方法。注意我们如何转移忽略'|'致船长:

Live On Coliru

qi::phrase_parse(
        f, l, *(qi::double_>>!qi::lit('%')) >> *(qi::double_>>'%'),
        qi::blank | '|', v, w);