2

我是 C++ 新手，在理解整个读取文件流的过程中遇到了一些麻烦。任何帮助将不胜感激...这就是我遇到问题的地方

我有一系列这样的结构；（不，我显然不允许使用字符串来存储这些东西，或者向量或我没有涉及的任何其他更高级的东西）......

struct Staff
{
    char title[TITLESIZE];
    char firstName[NAMESIZE];
    char familyName[NAMESIZE];
    char position[POSSIZE];
    char room[TITLESIZE];
    char email[POSSIZE];
};

然后我有一个这些结构的数组；

Staff record[MAXNOSTAFF];

数据包含在由制表符分隔的文本文件中。但是，某些字段可能包含空格。数据如下：

Dr Sherine ANTOUN Lecturer 4327 3.204 sherine_antoun@gmail.com

这是我在代码中写的...

//function prototypes
bool getRecord (ifstream& infile, Staff dataAr[], bool& fileFound);

int main()
{

    Staff record[MAXNOSTAFF];
    bool fileFound;
    ifstream infile;

    getRecord(infile, record, fileFound); //function call
    if (fileFound==true) 
    {
        cerr <<"Exiting Program"<<endl;
        exit(1);
    }

    return 0;
}

//function definitions
bool getRecord (ifstream& infile, Staff dataAr[], bool& fileFound)
{
    infile.open("phonebook.txt");

    if (infile)
    {
        fileFound = true;
        cout << "File " <<PHONEBOOK<< " opened successfully.\n\n";
    }
    else if (!infile)
    {
        fileFound =  false;
        cerr << "Error! File could not be opened. \n";
    }

    while (infile.good())
    {        

        for (int lineIndex=0; lineIndex<MAXNOSTAFF; lineIndex++)
            for (int titleIndex=0; titleIndex<TITLESIZE; titleIndex++)
            {
                cin.getline(dataAr[lineIndex].title[titleIndex], MAXNOSTAFF, '/t');
            }

    }
    //check it works properly
    for (int k=0;k<10; k++)
    {
        for (int m=0; m<11; m++)
        {
            cout << k <<". Title is : "<<dataAr[k].title[m]<<endl;
        }
    }    
    infile.close();
    return fileFound;
}

任何帮助将不胜感激..谢谢

4

2 回答 2

2

让我向您展示 Boost Spirit 方法来解析这样的输入数据。

如果你从一个像这样的结构开始

struct Staff
{
    std::string title;
    std::string firstName;
    std::string familyName;
    std::string position;
    std::string room;
    std::string email;
};

您可以使用 Spirit 语法，例如：

    column = lexeme [ *~char_("\t\r\n") ];
    start  = column >> '\t'  >> column >> '\t' >> column >> '\t' >> column >> '\t' >> column >> '\t' >> column;

并将所有行解析为一个向量，如：

    It f(std::cin), l;
    std::vector<Staff> staff_members;
    bool ok = qi::parse(f, l, grammar % qi::eol, staff_members);

    if (ok)
    {
        for(auto const& member : staff_members)
        {
            std::cout << boost::fusion::as_vector(member) << "\n";
        }
    } else
    {
        std::cout << "Parsing failed\n";
    }

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

这是完整的测试程序Live on Coliru，示例运行：

clang++ -std=c++11 -Os -Wall -pedantic main.cpp && ./a.out <<INPUT
Dr  Sherine ANTOUN  Lecturer    4327    3.204   sherine_antoun@gmail.com
Mr  Jason SCRYPT    Enthusiast  3472    9.204   jason_scrypt@yahoo.com
INPUT

输出：

(Dr Sherine ANTOUN Lecturer 4327 3.204 sherine_antoun@gmail.com)
(Mr Jason SCRYPT Enthusiast 3472 9.204 jason_scrypt@yahoo.com)
Remaining input '
'

完整列表

#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/tuple/tuple_io.hpp>

namespace qi = boost::spirit::qi;

struct Staff
{
    std::string title;
    std::string firstName;
    std::string familyName;
    std::string position;
    std::string room;
    std::string email;
};

BOOST_FUSION_ADAPT_STRUCT(Staff, 
    (std::string, title)
    (std::string, firstName)
    (std::string, familyName)
    (std::string, position)
    (std::string, room)
    (std::string, email))

template <typename It, typename Skipper = qi::unused_type>
    struct grammar : qi::grammar<It, Staff(), Skipper>
{
    grammar() : grammar::base_type(start)
    {
        using namespace qi;
        column = lexeme [ *~char_("\t\r\n") ];
        start  = column >> '\t'  >> column >> '\t' >> column >> '\t' >> column >> '\t' >> column >> '\t' >> column;
    }
    private:
    qi::rule<It, std::string(), Skipper> column;
    qi::rule<It, Staff(), Skipper> start;
};

int main()
{
    std::cin.unsetf(std::ios::skipws);

    typedef boost::spirit::istream_iterator It;
    grammar<It> grammar;

    It f(std::cin), l;
    std::vector<Staff> staff_members;
    bool ok = qi::parse(f, l, grammar % qi::eol, staff_members);

    if (ok)
    {
        for(auto const& member : staff_members)
        {
            std::cout << boost::fusion::as_vector(member) << "\n";
        }
    } else
    {
        std::cout << "Parsing failed\n";
    }

    if (f != l)
        std::cout << "Remaining input '" << std::string(f, l) << "'\n";
}
于 2014-03-18T15:19:43.800 回答
1

由于您不能使用std::stringand std::vector，sscanf()因此可能是您的选择：

    while (infile.good())
    {
        char line[BUF_SIZE];
        for (int lineIndex=0; lineIndex<MAXNOSTAFF; lineIndex++)
        {
            infile.getline(line, BUF_SIZE);
            sscanf(line, "%s %s %s %[^\t] %s %s", dataAr[lineIndex].title, dataAr[lineIndex].firstName, dataAr[lineIndex].familyName, dataAr[lineIndex].position, dataAr[lineIndex].room, dataAr[lineIndex].email);
        }
    }

请注意%[^\t]格式说明符，它将匹配每个不是的字符\t（因为 ^），以便可以正确读取包含空格的文件。我不知道究竟哪些字段包含空格，所以我只是写一个例子。
编辑：
如果允许使用，您可以在从文件流中获取一行后拆分字符串std::string：std::stirngstream

while (infile.good())
    {
        char line[BUF_SIZE];
        for (int lineIndex=0; lineIndex<MAXNOSTAFF; lineIndex++)
        {
            infile.getline(line, BUF_SIZE);
            stringstream ss(line);
            std::string s;
            getline(ss, s, '\t'); // get the first field

            getline(ss, s, '\t'); // get the second field

            // ...
        }
    }
于 2014-03-17T01:39:00.400 回答