編程不能找到我的文件

Question

我這樣做，我需要一個編程準備這是名group1.txt和group2.txt兩個獨立的輸入文件，然後創建一個發現平均得分爲每個組

一個編程我寫了這個，但它不能找到我的文件，原因我不知道 一些幫助，也許謝謝！

下面是代碼：

#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; 
} 

只需要知道如何讓我的檔案！我把桌面，我的文檔，項目，與C + +文件，我不知道如何！有些我需要有一個軟拷貝，我知道它是否能夠在那裏找到我的文件！

Answer 1

您指定文件名的方式爲「group1.txt」，只有程序在編譯程序的當前工作目錄中才能找到該文件。

有兩種方法來解決這個問題：

• 實際上將文件複製到該程序的工作目錄。在許多情況下，這將是可執行文件所在的目錄。
• 使用絕對文件名，例如Win7系統上存儲在桌面上的文件（「c：\ Users \ youruser \ Desktop \ group1.txt」）