Open files by passing variable as file name

Question

I have the following situation, a file "tables.txt" where I have around 2000 lines with filename to which I want to pass row by row taking the name of the files and check which ones exist and which ones not.

The problem occurs when after catching the file name to be checked try to pass it to open it later ifstream obj (arq), the strange thing is that if I type passing the string with the path and not the variable arq that is string the file is read smoothly.

Follows the code used:

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


int main () {
    string line;    
    ifstream myfile ("tabelas.txt"); 
    if (myfile.is_open()){
        while (! myfile.eof() ) 
        {
            getline (myfile,line); 
            string arq = "..\\dados\\"+ line+".csv";

            ifstream obj (arq);
            if (!obj) 
                cerr<<"Ocorreu um erro ao abrir o arquivo"<<endl;
        }
    myfile.close();
    }

    else cout << "Unable to open file"; 

    return 0;
}

Follows image demonstrating error:

http://prntscr.com/by4opb

Answer 1

I rebuilt it to read files the way I normally do.

Follow the code below:

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

int main () {
    ifstream myfile ("x.txt", ifstream::in);
    string line;
    if(myfile.is_open()){
        while(getline(myfile, line)){
            line = "..\\dados\\" + line + ".csv";
            ifstream testfile (line.c_str(), ifstream::in);
            if(testfile.is_open()){
                cout << "O arquivo '"<< line <<"' existe\n";
            }else{
                cout << "O arquivo '" << line << "' nao existe\n";
            }
        }
    }
    myfile.close();


    return 0;
}