Question

我正在编程，要求我准备两个单独的输入文件，名称为group1.txt和group2.txt，然后创建一个程序，找到每个组的平均分数

我写了这个，但由于我不知道的原因，它无法找到我的文件一些帮助也许谢谢！

以下是代码：

#include <iostream>
#include <fstream>
#include <string>
#include <stdlib.h>
using namespace std;

int main()

{
  string line;
  const int num_lines = 10; //change numline to any number you like, its to set the       size of the array
  string sub_code[num_lines];
  float avrgs[num_lines];
  string sub_code2[num_lines];
  float avrgs2[num_lines];
  ifstream myfile ("group1.txt");
  int sum=0,score=0,j=1,i=0,i2=0;
  double ave=0.0;


  if (myfile.is_open())
  {
    while ( myfile>>line )
    {
    //cout<<line<<endl;
    sub_code[i] = line;
     while ( myfile>>score && score <100 && score >= 0)
    {
    sum += score;
     j++;
    }
    ave = sum/j;
    j=1;
    sum = 0;
    avrgs[i]=ave;
    i++;
     }
  }
   else
   {cout << "Unable to open file"<<endl;}
myfile.close();

//this is for the secode line
ifstream myfile2 ("group2.txt");
if (myfile2.is_open())
 {
    while ( myfile2>>line )
    {
    //cout<<line<<endl;
    sub_code2[i2] = line; //add all the subject code into the array to store sub codes
    while ( myfile2>>score && score <100 && score >= 0) //well basically the score    should be between 0 - 100,
    {                                                   //so -999 wont be read. Can      change to while ( myfile2>>score && score!=-999)
sum += score;                                       //read each grade and add it to sum
j++;                                                //just to know how many grades are there so that division can be done
}
ave = sum/j;                                        //find the average.
j=1;                                                //set j back to 1, cause j is used to count the number of marks.
sum = 0;                                            //since its sum+=score, we need to set sum back to 0, or else it will be adding on to the old marks
avrgs2[i2]=ave;                                     //add that calculated ave into the array for average
i2++;                                               //i2 is to basically know how many entries are in the file for grp2
}
  }
      else
      {cout << "Unable to open file"<<endl;}
    myfile2.close();

int gr1,gr2;
//outputing the averages and so on...
for (gr1=0;gr1<i;gr1++)
{
for(gr2=0;gr2<i2;gr2++)
{
    if(sub_code[gr1]==sub_code2[gr2]) // compare subject id before displaying
    {
    cout<<sub_code[gr1]<<"\t"<<" 1 "<<avrgs[gr1]<<endl;
    cout<<sub_code2[gr2]<<"\t"<<" 2 "<<avrgs2[gr2]<<endl;
    break;
    }
}
}
//
cout<<endl;

double grpave1=0,grpave2=0;
//to find the average of each group
for (gr1=0;gr1<i;gr1++)
{
grpave1+=avrgs[gr1];
}
for(gr2=0;gr2<i2;gr2++)
{
grpave2+=avrgs2[gr2];
}
grpave1=grpave1/i;
grpave2=grpave2/i2;
cout<<"Average for group 1:"<<grpave1<<endl;
cout<<"Average for group 2:"<<grpave2<<endl;

system("pause");

return 0;
}

只需要知道如何获取我的文件！我已经放入了桌面，我的文档，项目，使用C ++文件，我不知道如何！我需要有一份软拷贝，我知道它是否能在那里找到我的文件！

Answer 1

指定文件名的方式为“group1.txt”，只有程序在编译程序的当前工作目录中才能找到该文件。

有两种方法可以解决这个问题：

实际上将文件复制到程序的工作目录中。在许多情况下，这将是可执行文件所在的目录。
使用绝对文件名，例如（“c：\ Users \ youruser \ Desktop \ group1.txt”表示存储在Win7系统桌面上的文件。

程序设计无法找到我的文件

1 个答案: