#include <iostream>
#include <iomanip>
#include <fstream>
#include <string>
using namespace std;
struct subscriberName
{
string first;
string last;
int custID;
};
struct address
{
string address2;
string city;
string state;
int zipcode;
};
struct date
{
string month;
int day;
int year;
};
struct renewal_information
{
int monthsLeft;
date da;
};
struct subscriberInfo
{
subscriberName si;
address ad;
renewal_information ri;
};
int main()
{
void OpenFileIn(ifstream& FileIn, string& FilenameIn);
void OpenFileOut(ofstream& FileOut, string& FilenameOut);
bool ProcessCustInfo(bool& truth, ifstream& FileIn);
void OutputCustInfo(ifstream& FileIn, ofstream& FileOut);
ifstream FileIn;
ofstream FileOut;
string FilenameIn;
string FilenameOut;
bool truth;
subscriberInfo si;
OpenFileIn(FileIn, FilenameIn);
OpenFileOut(FileOut, FilenameOut);
ProcessCustInfo(truth, FileIn);
OutputCustInfo(FileIn, FileOut);
return 0;
}
bool ProcessCustInfo(bool& truth, ifstream& FileIn, subscriberInfo& si)
{
getline(FileIn, si.sn.first, '\n'); //here
getline(FileIn, si.sn.last, '\n');
getline(FileIn, si.sn.custID, '\n');
getline(FileIn, si.ad.address2, '\n');
getline(FileIn, si.ad.city, '\n');
getline(FileIn, si.ad.state, '\n');
getline(FileIn, si.ad.zipcode, '\n');
getline(FileIn, si.ri.monthsLeft '\n'); //to here
}
void OutputCustInfo(ifstream& FileIn, ofstream& FileOut, subscriberInfo& si)
{
if(si.ri.monthsLeft=0) //here down to
{
FileOut << string(55,'*') << endl;
FileOut << si.sn.first << " " << si.sn.last << "(" << si.sn.custID << ")" << endl;
FileOut << sn.ad.address2 << endl;
FileOut << sn.ad.city << ", " << sn.ad.state <<sn.ad.zipcode << endl;
FileOut << "The last renewal notice was sent on " <<sn.ri.da.month << " " << sn.ri.da.day << ", " << sn.ri.da.year << endl; //here
FileOut << string(55,'*') << endl;
}
}
我找不出是什么原因导致此错误。它发生在所有getline调用都在的第一个函数中。编译器专门调用第三个,第五个和最后一个,但我很确定它们都有问题。收到错误“没有匹配函数调用'getline(std :: ifstream&,int&,char)'”
getline(FileIn, si.sn.custID, '\n');
这是一个问题:
'geline()'只接受'串'作为第二个参数,你需要把它放在一个字符串,然后将其转换成'int'或其他类型。由于结构中的某些成员的类型为'int' –
@JohnMarkCaguicla将成员更改为字符串会更容易吗?这可能导致什么问题? – LCro
这取决于你将如何使用它们,如果你只是显示它们,可以将它们改为'string',但为了更实际的目的,你可以使用'custID'作为int '传递给一个函数。在这种情况下,只需将其作为“字符串”进行检索,然后将其转换为“int”。 –