How can i open multiple files in one run in C++? - c++

Question

Hey I have several Files that I want to read. The user would input what file they want to open on console. And if they want they can change and read another file, they would do so.
This is how i went by doing (this is just rough copy, the code i have is too big will take too long to understand)
#include <string>
#include <iostream>
#include <vector>
#include <fstream>
#include <iomanip>
#include<algorithm>
using namespace std;
int main()
{
string FileName;
string Line;
cout << "Input File Directory To Open :" << endl;
cin >> FileName;
ifstream File;
File.open(FileName);
string Input;
do {
cout << "Enter R to Read Display Data In File Or C to Read Another File Or X To Exit" << endl;
cin >> Input;
if (Input == "R")
{
while (getline(File, Line))
{
cout << Line;
}
}
else if (Input == "C")
{
string FileName2;
cout << "Enter New File Directory To Open: " << endl;
cin >> FileName2;
//? ? ? ?
}
} while (Input != "X");
}
Since it's in do while loop, when user input C and input directory to read another file, so that they can cout new file when they input R next.
My question is how should i overwrite the FileName2 with FileName?
Hope It makes sense Thank you

Answer 1

First close the file that is already open:
File.close();
Then open the new one:
File.open(FileName2);

Answer 2

Both answers keep your idea of an open file handle. However, a better approach is to use RAII. In that case you only register the file name and only open the file when you need it.
int main()
{
std::cout << "Input File Directory To Open :\n";
std::string fileName;
std::cin >> fileName;
while (true) {
std::cout << "Enter R to Read Display Data In File Or C to Read Another File Or X To Exit\n";
std::string input;
std::cin >> input;
if (input == "X") break;
if (input == "R") {
std::ifstream file(filename);
std::string line;
while (std::getline(file, line)) {
std::cout << line;
}
// due to RAII the file will close when the object goes out of scope
} else if (input == "C") {
std::cout << "Enter New File Directory To Open: \n";
std::cin >> fileName;
}
}
}

Answer 3

You don't need a new variable. Simply assign the value to the existing one, close the currently opened file, and then open the new file inside the loop:
...
else if (Input == "C")
{
cout << "Enter New File Directory To Open: " << endl;
cin >> FileName;
File.close();
File.open(FileName);
}
...