我是c++新手,理解整个读取文件流的事情有点麻烦。任何帮助都会很感激……这就是我遇到麻烦的地方
我有一个像这样的结构数组;(不,我不允许使用字符串来存储这些东西,或向量或任何其他更高级的东西,我没有涉及)…
struct Staff
{
char title[TITLESIZE];
char firstName[NAMESIZE];
char familyName[NAMESIZE];
char position[POSSIZE];
char room[TITLESIZE];
char email[POSSIZE];
};
然后我有一个这样结构的数组;
Staff record[MAXNOSTAFF];
数据包含在以制表符分隔的文本文件中。但是,有些字段可能包含空格。数据如下:
Dr Sherine ANTOUN Lecturer 4327 3.204 sherine_antoun@gmail.com
这是我在我的代码中写的…
//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.nn";
}
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;
}
任何帮助都将是非常感激的。谢谢你 让我向您展示Boost Spirit解析输入数据的方法。
如果以像
这样的结构体开头struct Staff
{
std::string title;
std::string firstName;
std::string familyName;
std::string position;
std::string room;
std::string email;
};
你可以使用Spirit语法:
column = lexeme [ *~char_("trn") ];
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 failedn";
}
if (f != l)
std::cout << "Remaining input '" << std::string(f, l) << "'n";
下面是完整的测试程序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 '
'
完整清单
#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_("trn") ];
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 failedn";
}
if (f != l)
std::cout << "Remaining input '" << std::string(f, l) << "'n";
}
既然你不能使用std::string
和std::vector
, sscanf()
可能是你的选择:
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);
}
}
注意%[^t]
格式说明符,它将匹配每个不是t
的字符(因为^),因此包含空格的字段可以被正确读取。我不知道哪些字段包含空格,所以我只写一个例子。
编辑:
如果允许使用std::string
和std::stirngstream
,则可以在从文件流中获取一行后拆分字符串:
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
// ...
}
}