Question

我正在尝试逐行读取文本文件并将数据提取到我的程序中。从本质上讲，C ++程序将读取已经制作的输入文件，格式为：

     Republican Senator John McMahon
     Democrat Mayor Steven Markel
     Republican Judge Matt Stevens
     Democrat Senator Anthony Quizitano
     S R
     M D
     J R
     ...
     ..
     ..

格式基本上是前3行包括Party，Position和Name，接下来的行表示“结果”，其形式为：

[第一封信] [投票选举]

所以，例如，如果你看到S R，这意味着参议员有1票，而且是共和党候选人。

这是我到目前为止所拥有的：

            #include<fstream>
            int main()
            {
            std::ifstream input("file.txt");
            }

这是我的理解，这将允许我输入文件，并逐行进行，但我不知道如何从这里开始实现这个...任何帮助？

谢谢！

Answer 1

这是为了娱乐和荣耀，一种基于Boost Spirit的实现。我添加了更多假投票输入，只是因此可能会显示一些内容。

我不确定候选人和投票之间是否存在1：1的关系（我不是美国公民，我不知道列出的候选人是投票还是投票）。所以我决定只使用假数据。

const std::string input = 
    "Republican Senator John McMahon\n"
    "Democrat Senator Anthony Quizitano\n"
    "S R\n"
    "S R\n"
    "S R\n"
    "Democrat Mayor Steven Markel\n"
    "Republican Judge Matt Stevens\n"
    "M D\n"
    "J R\n"
    "S R\n"
    "S R\n";

然而，该代码可用于两种目的。

我对输入的显示顺序不重要。
但是，您可以断言，单个字母（S，M，J）实际上对应于该点之前列出的位置。通过使用posletter_check

在 http://liveworkspace.org/code/d9e39c19674fbf7b2419ff88a642dc38

上观看演示

#define BOOST_SPIRIT_USE_PHOENIX_V3
#define BOOST_RESULT_OF_USE_DECLTYPE
#include <boost/fusion/adapted.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <iomanip>

namespace qi    = boost::spirit::qi;
namespace phx   = boost::phoenix;

struct Candidate { std::string party, position, name; };

BOOST_FUSION_ADAPT_STRUCT(Candidate, (std::string, party)(std::string, position)(std::string, name))

typedef std::map<std::pair<char, char>, size_t> Votes;
typedef std::vector<Candidate> Candidates;

template <typename It>
    struct parser : qi::grammar<It>
{
    mutable Votes _votes;
    mutable Candidates _candidates;

    parser() : parser::base_type(start)
    {
        using namespace qi;
        using phx::bind; using phx::ref; using phx::val;

        start = (line % eol) >> *eol >> eoi;

        line = 
              vote      [ phx::bind(&parser::register_vote, phx::ref(*this), _1) ]
            | candidate [ phx::push_back(phx::ref(_candidates), _1) ]
            ;

        vote %= graph   
                        // Comment the following line to accept any single
                        // letter, even if a matching position wasn't seen
                        // before:
                        [ _pass = phx::bind(&parser::posletter_check, phx::ref(*this), _1) ]  
            >> ' ' 
            >> char_("RD")
            ;

        candidate = (string("Republican") | string("Democrat"))
            >> ' ' 
            >> as_string [ +graph ]
            >> ' ' 
            >> as_string [ +(char_ - eol) ]
            ;
    }

  private:
    bool posletter_check(char posletter) const
    {
        for (auto& c : _candidates)
            if (posletter == c.position[0])
                return true;
        return false;
    }
    void register_vote(Votes::key_type const& key) const
    {
        auto it = _votes.find(key);
        if (_votes.end()==it)
            _votes[key] = 1;
        else
            it->second++;
    }

    qi::rule<It, Votes::key_type()> vote;
    qi::rule<It, Candidate()> candidate;
    qi::rule<It> start, line;
};

int main()
{
    const std::string input = 
        "Republican Senator John McMahon\n"
        "Democrat Senator Anthony Quizitano\n"
        "S R\n"
        "S R\n"
        "S R\n"
        "Democrat Mayor Steven Markel\n"
        "Republican Judge Matt Stevens\n"
        "M D\n"
        "J R\n"
        "S R\n"
        "S R\n";

    std::string::const_iterator f(std::begin(input)), l(std::end(input));

    parser<std::string::const_iterator> p;

    try
    {
        bool ok = qi::parse(f,l,p);
        if (ok)   
        {
            std::cout << "\ncandidate list\n";
            std::cout << "------------------------------------------------\n";
            for (auto& c : p._candidates)
                std::cout << std::setw(20) << c.name << " (" << c.position << " for the " << c.party << "s)\n";

            std::cout << "\nVote distribution:\n";
            std::cout << "------------------------------------------------\n";
            for (auto& v : p._votes)
                std::cout << '(' << v.first.first << "," << v.first.second << "): " << v.second << " votes " << std::string(v.second, '*') << "\n";
        }
        else std::cerr << "parse failed: '" << std::string(f,l) << "'\n";

        if (f!=l) std::cerr << "trailing unparsed: '" << std::string(f,l) << "'\n";
    } catch(const qi::expectation_failure<std::string::const_iterator>& e)
    {
        std::string frag(e.first, e.last);
        std::cerr << e.what() << "'" << frag << "'\n";
    }
}

输出：

candidate list
------------------------------------------------
        John McMahon (Senator for the Republicans)
   Anthony Quizitano (Senator for the Democrats)
       Steven Markel (Mayor for the Democrats)
        Matt Stevens (Judge for the Republicans)

Vote distribution:
------------------------------------------------
(J,R): 1 votes *
(M,D): 1 votes *
(S,R): 5 votes *****

从文本文件中读取输入并将其解压缩 - C ++

1 个答案: