在c ++中读取文本文件到数组的输入

-1 c++ arrays fstream getline ifstream

好吧,温柔,因为我非常喜欢编程.到目前为止,我只学习了C++,并且我正在运行Visual Studio 2010作为我的编译器.对于这个程序,我试图从文本输入文件中读取并将信息写入一组三个数组.一个阵列将处理名称列表,另外两个阵列分别用于工作小时数和小时工资率.我将使用后两者来计算一组收入,并将整个事物输出到另一个文本文件作为报告.然而,我的问题是获取第一个数组的输入.我正在使用的输入文件的文本排列如下:

J. Doe*35 12.50 J. Dawn*20 10.00 .........

由于我试图使用ifstream getline来获取星号充当分隔符的名称,并将以下两个数字写入另外两个数组,因此这些名称会被星号所拖尾.后两个值由空格分隔,所以我认为它们不会引起任何问题.我确定还有其他错误需要处理,但我需要先处理第一个错误才能开始调试其余的错误.

我的问题出现在我调用inFile.getline的行,因为我收到以下错误:错误C2664:'std :: basic_istream <_Elem,_Traits>&std :: basic_istream <_Elem,_Traits> :: getline(_Elem*,std :: streamsize,_Elem)':无法将参数1从'std :: string'转换为'char*'.

从我在其他地方读到的,(我认为)问题源于尝试将字符串写入char数组,由于它们具有不同的数据类型,因此无法工作.我不确定是否存在其他可行的方法来获取名称,因为我需要使用分隔符将名称与数值分开.如何解决这个问题的任何建议将不胜感激.

这是我写的来源:

#include <iostream>
#include <fstream>
#include <iomanip>
#include <string>
using namespace std;

const int EMP_NUM = 5;
const int BASE_HOURS = 40;
const char N_SIZE = 8;

int main()
{
 int i;
 double rEarnings, oEarnings, tEarnings,
 trEarnings, toEarnings, ttEarnings;
 ifstream inFile;
 ofstream outFile;
 inFile.open("records.txt");
 outFile.open("report.txt");

 outFile << setprecision(2) << showpoint << fixed;

 outFile << setw(50) << "Payroll Report" << "\n\n";
 outFile << "EMPLOYEE NAME" << setw(25) << "REGULAR EARNINGS" << setw(25) << "OVERTIME EARNINGS" << setw(25) << "TOTAL EARNINGS" << endl;

 string nameAr[EMP_NUM];
 int hoursAr[EMP_NUM];
 double hrateAr[EMP_NUM];

 for (int i = 0; i < EMP_NUM; i++) // Get input from our input file.
 {
  inFile.getline(nameAr[i], EMP_NUM, "*");
  inFile >> hoursAr[i] >> hrateAr[i];
 }

 for (int i = 0; i < EMP_NUM; i++) // Make the calculations to be sent to our report.
 {
  char nameAr[N_SIZE];
  int hoursAr[N_SIZE];
  double hrateAr[N_SIZE];

  if (hoursAr[i] > 40) // For employees with overtime hours.
  {
  // double rEarnings, double oEarnings, double tEarnings,
  // double trEarnings, double toEarnings, double ttEarnings;
  // rEarnings = 0, oEarnings = 0, tEarnings = 0,
  // trEarnings = 0, toEarnings = 0, ttEarnings = 0;

   rEarnings = BASE_HOURS * hrateAr[i];
   oEarnings = (hoursAr[i] - BASE_HOURS) * hrateAr[i] * 1.5;
   tEarnings = rEarnings + oEarnings;
   trEarnings += rEarnings;
   toEarnings += oEarnings;
   ttEarnings += tEarnings;
   outFile << left << nameAr[i];
   // << setw(25) << right << rEarnings << setw(25) << right << oEarnings << setw(25) << right << tEarnings << endl;

  } 
  else // For employees without overtime hours.
  {
  // double rEarnings, double oEarnings, double tEarnings,
  // double trEarnings, double toEarnings, double ttEarnings;
  // rEarnings = 0, oEarnings = 0, tEarnings = 0,
  // trEarnings = 0, toEarnings = 0, ttEarnings = 0;

   rEarnings = hoursAr[i] * hrateAr[i];
   oEarnings = 0;
   tEarnings = rEarnings + oEarnings;
   trEarnings += rEarnings;
   toEarnings += oEarnings;
   ttEarnings += tEarnings;
   outFile << left << nameAr[i]; 
   // << setw(25) << right << rEarnings << setw(25) << right << oEarnings << setw(25) << right << tEarnings << endl;
  }
 }

 outFile << endl << endl;

 outFile << setw(33) << trEarnings << " *" << setw(23) << toEarnings << " *" << setw(23) << ttEarnings << " *\n\n";

 outFile << left << "TOTAL EMPLOYEES" << " " << (i - 1);

 inFile.close(); outFile.close();

 return 0;
}
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我已经包含了整个程序,让您了解我打算在哪里编写代码.在此先感谢您的帮助!

wkl*_*wkl 5

fstreamgetline功能有这些签名.

std::istream& getline (char* s, streamsize n );
std::istream& getline (char* s, streamsize n, char delim );
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在你的代码中......

string nameAr[EMP_NUM];
// stuff...
inFile.getline(nameAr[i], EMP_NUM, "*");
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您正在尝试使用std::string这里getline想要一个char*-你不能从转换stringchar*这个样子.

一种方法是你可以使用字符缓冲区来存储内容getline,就像这样.

const int BUFFER_SIZE = 1024;
char buffer[BUFFER_SIZE];
inFile.getline(buffer, BUFFER_SIZE, '*');
// more stuff...
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更好的解决方案

我建议使用std::getlinefrom <string>,因为你可以使用string类型而不是C字符串.

#include <string>

// stuff...
std::getline(inFile, nameAr[i], '*');
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在您学习之后编辑链接