如何使用boost :: xpressive在语义动作中填充带有结构的向量

时间:2018-02-22 01:24:15

标签: boost boost-xpressive

每次检测到匹配时,我都试图将数据结构插入到向量中,但即使在编译时也是如此。代码是下一个:

#include <string>
#include <boost/xpressive/xpressive.hpp>
#include <boost/xpressive/regex_actions.hpp>

using namespace boost::xpressive;

struct Data
{
    int integer;
    double real;
    std::string str;

    Data(const int _integer, const double _real, const std::string& _str) : integer(_integer), real(_real), str(_str) { }
};

int main()
{
    std::vector<Data> container;

    std::string input = "Int: 0 - Real: 18.8 - Str: ABC-1005\nInt: 0 - Real: 21.3 - Str: BCD-1006\n";

    sregex parser = ("Int: " >> (s1 = _d) >> " - Real: " >> (s2 = (repeat<1,2>(_d) >> '.' >> _d)) >> " - Str: " >> (s3 = +set[alnum | '-']) >> _n)
                    [::ref(container)->*push_back(Data(as<int>(s1), as<double>(s2), s3))];

    sregex_iterator cur(input.begin(), input.end(), parser);
    sregex_iterator end;

    for(; cur != end; ++cur)
        smatch const &what = *cur;

    return 0;
}

编译失败&#34; push_back&#34;由于我在内部使用Data对象而导致语义动作,并且它无法懒惰地使用它(我猜,我不太确定)。

拜托,有人可以帮我吗?

注意 - 我不熟悉MS VS 2010(不完全符合c ++ 11标准),所以请不要使用可变参数模板和emplace_back解决方案。谢谢。

1 个答案:

答案 0 :(得分:3)

使用Xpressive

你应该让这个动作成为一个懒惰的演员。您的Data构造函数调用不是。

<强> Live On Coliru

#include <string>
#include <boost/xpressive/xpressive.hpp>
#include <boost/xpressive/regex_actions.hpp>

namespace bex = boost::xpressive;

struct Data {
    int integer;
    double real;
    std::string str;

    Data(int integer, double real, std::string str) : integer(integer), real(real), str(str) { }
};

#include <iostream>

int main() {
    std::vector<Data> container;

    std::string const& input = "Int: 0 - Real: 18.8 - Str: ABC-1005\nInt: 0 - Real: 21.3 - Str: BCD-1006\n";

    using namespace bex;
    bex::sregex const parser = ("Int: " >> (s1 = _d) >> " - Real: " >> (s2 = (repeat<1,2>(_d) >> '.' >> _d)) >> " - Str: " >> (s3 = +set[alnum | '-']) >> _n)
        [bex::ref(container)->*bex::push_back(bex::construct<Data>(as<int>(s1), as<double>(s2), s3))];

    bex::sregex_iterator cur(input.begin(), input.end(), parser), end;

    for (auto const& what : boost::make_iterator_range(cur, end)) {
        std::cout << what.str() << "\n";
    }

    for(auto& r : container) {
        std::cout << "[ " << r.integer << "; " << r.real << "; " << r.str << " ]\n";
    }
}

打印

Int: 0 - Real: 18.8 - Str: ABC-1005

Int: 0 - Real: 21.3 - Str: BCD-1006

[ 0; 18.8; ABC-1005 ]
[ 0; 21.3; BCD-1006 ]

使用精神

我会用精神来做这件事。 Spirit具有直接解析为基础数据类型的原语,这样不易出错且效率更高。

Spirit Qi(V2)

使用Phoenix,非常相似: Live On Coliru

使用Fusion适应,它变得更有趣,更简单:

<强> Live On Coliru

现在想象一下:

  • 您希望匹配不区分大小写的关键字
  • 你想让空白无关紧要
  • 您想接受空行,但不接受
  • 之间的随机数据

你会如何在Xpressive中做到这一点?以下是你如何使用Spirit做到的。请注意,附加约束如何不会改变语法。与基于正则表达式的解析器形成对比。

<强> Live On Coliru

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

struct Data {
    int integer;
    double real;
    std::string str;
};

BOOST_FUSION_ADAPT_STRUCT(Data, integer, real, str);

#include <iostream>

int main() {
    std::vector<Data> container;
    using It = std::string::const_iterator;

    std::string const& input = "iNT: 0 - Real: 18.8 - Str: ABC-1005\n\nInt: 1-Real:21.3 -sTR:BCD-1006\n\n";

    qi::rule<It, Data(), qi::blank_type> parser = qi::no_case[
            qi::lit("int") >> ':' >> qi::auto_ >> '-' 
            >> "real" >> ':' >> qi::auto_ >> '-' 
            >> "str" >> ':' >> +(qi::alnum|qi::char_('-')) >> +qi::eol
        ];

    It f = input.begin(), l = input.end();
    if (parse(f, l, qi::skip(qi::blank)[*parser], container)) {
        std::cout << "Parsed:\n";
        for(auto& r : container) {
            std::cout << "[ " << r.integer << "; " << r.real << "; " << r.str << " ]\n";
        }
    } else {
        std::cout << "Parse failed\n";
    }

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

仍打印

Parsed:
[ 0; 18.8; ABC-1005 ]
[ 1; 21.3; BCD-1006 ]

进一步的想法:你将如何

  • 解析科学记数法?负数?
  • 正确解析十进制数(假设您实际上正在解析财务金额,您可能不希望不精确的浮点表示)

精神X3

如果你可以使用c ++ 14,Spirit X3可以更高效,并且比Spirit Qi或Xpressive方法编译更快

<强> Live On Coliru

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

struct Data {
    int integer;
    double real;
    std::string str;
};

BOOST_FUSION_ADAPT_STRUCT(Data, integer, real, str);

namespace Parsers {
    using namespace boost::spirit::x3;

    static auto const data 
        = rule<struct Data_, ::Data> {} 
        = no_case[
            lit("int") >> ':' >> int_ >> '-' 
            >> "real" >> ':' >> double_ >> '-' 
            >> "str" >> ':' >> +(alnum|char_('-')) >> +eol
        ];

    static auto const datas = skip(blank)[*data];
}

#include <iostream>

int main() {
    std::vector<Data> container;

    std::string const& input = "iNT: 0 - Real: 18.8 - Str: ABC-1005\n\nInt: 1-Real:21.3 -sTR:BCD-1006\n\n";

    auto f = input.begin(), l = input.end();
    if (parse(f, l, Parsers::datas, container)) {
        std::cout << "Parsed:\n";
        for(auto& r : container) {
            std::cout << "[ " << r.integer << "; " << r.real << "; " << r.str << " ]\n";
        }
    } else {
        std::cout << "Parse failed\n";
    }

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

打印(它变得无聊):

Parsed:
[ 0; 18.8; ABC-1005 ]
[ 1; 21.3; BCD-1006 ]