Very Easy Question

Code:

ifstream file_input("myinputfile.txt");

cout << file_input.something; // prints "myinputfile.txt"

What is the "something" above?

Like this?

Code:

#include <iostream>
#include <string>

struct ifstream
{
    std::string something;
    ifstream(const std::string& sth): something(sth) {}
};

using std::cout;

int main(){
    ifstream file_input("myinputfile.txt");

    cout << file_input.something; // prints "myinputfile.txt"
}

Or what is your question? I don't think ifstream stores the name of the file if that's what you mean. If you opened the file, the name was available to you - why can't you use that?

Originally Posted by anon

Like this?

Code:

#include <iostream>
#include <string>

struct ifstream
{
    std::string something;
    ifstream(const std::string& sth): something(sth) {}
};

using std::cout;

int main(){
    ifstream file_input("myinputfile.txt");

    cout << file_input.something; // prints "myinputfile.txt"
}

Or what is your question? I don't think ifstream stores the name of the file if that's what you mean. If you opened the file, the name was available to you - why can't you use that?

I want to have an error when the file can't be opened. I want it to say:

cout << "Error! " << (filename) << " could not be opened!";

The name of the file will eventually be a command line argument to the program, therefore it will change.

Not possible?

If it is a command line argument then it is stored in a variable? Why can't you use that?

Code:

string name = "somefile.txt";
ifstream fin(name.c_str());
if (!fin) {
    cout << "Couldn't open " << name;
}

You're specifying the filename when you open the file, so you must know it.
If it can't be opened, just print that filename.

Originally Posted by Elysia

You're specifying the filename when you open the file, so you must know it.
If it can't be opened, just print that filename.

Ah yes... Nevermind

When I (eventually) code it to work with command line arguments, I can access it that way through those arguments

Thanks