simple I/O question

Hi everyone!

I need this: the user can choose the name of the output file, however, if the file can not be open, the program will automatically use cout. I don´t want to do this with if, because there are many regions in the program that use this output file, the code will become dirty if I use if

A sample high-level algorithm would be:

Code:

out_file.open(user_file);
if(!out_file){
   out_file = cout;
}

//use out_file at will

Thanks any help!

I'm afraid I haven't worked with C++-specific stuff in a very long time, so I can only offer a C-style solution, but it'll still work in C++, and the idea might even be applicable to C++ objects.

In C, a lot of the standard output functions are simply duplicates of file-handling functions with stdout specified as the file pointer. You could have a global variable be a file pointer. Set it equal to stdout if the user-specified file cannot be opened, and then use that file pointer in the file-handling functions.

You can redirect output from cout to a file if you successfully open the file. If the file is not open you write to cout as normal.

Code:

#include <iostream>
#include <fstream>

using namespace std;

int main()
{
    ofstream output("output.txt");
    streambuf* saved_cout = NULL;

    // If output is open, make cout actually write to output
    if( output.is_open() )
    {
        saved_cout = cout.rdbuf();  // Save existing cout for later
        cout.rdbuf(output.rdbuf()); // Redirect cout to use file
    }

    // This will go to cout if file was not opened, otherwise it is sent to the file
    cout << "Hello World!" << endl;

    // Restore cout if it had been redirected
    if( saved_cout ) cout.rdbuf(saved_cout);

    // Output message to console per normal cout operations.
    cout << "Goodbye!" << endl;

}

This may or may not help you out.

Thanks for both reply, really good ideas. sean_mackrory your idea is very simple and clean, my fault that I didn´t think this solution . I´ve changed your code hk_mp5kpdw, hope you don´t mind.

Code:

#include <iostream>
#include <fstream>

using namespace std;

int main()
{
    fstream output("output.txt");

    // Presume that the file was correctly open
    ostream out(output.rdbuf());

    // If not, redirect do cout
    if( !output )
    {
        out.rdbuf(cout.rdbuf());
    }

    // This will go to cout if file was not opened, otherwise it is sent to the file
    out << "Hello World!" << endl;

}

that solution still doesnt meet your requirments though. because

I need this: the user can choose the name of the output file,

and in the code you have

Code:

fstream output("output.txt");

so the user has no saying on choosing the filename as the file name is coded in the program.

atleast thats what i assume from your post, correct me if im wrong.

Originally Posted by InvariantLoop

that solution still doesnt meet your requirments though. because

...

and in the code you have

...

so the user has no saying on choosing the filename as the file name is coded in the program.

atleast thats what i assume from your post, correct me if im wrong.

At least for my post, it was just an example of how to redirect output streams. What I posted does not address the issue of a user choosing the name of the file nor was it intended to. I would guess the OP's response was simply more experimentation with the concept of redirection, i.e. playing around with the concept.

hk_mp5kpdw yes i know that my reply was towards the orignal poster of this thread, not you hehe.

im facing exactly the same problem and i have yet to find a solution

Unless I'm totally drunk std::cout is an ostream as well as std::fstream.

Code:

void WriteOrWhatever(std::ostream& stream)
{
   stream << "Stuff!";
}

void SelectFile(const std::string& filename)
{
   std::ofstream file(filename.c_str());

   if(file.fail())
   {
      WriteOrWhatever(std::cout);
   }
   else
   {
      WriteOrWhatever(file);
      file.close();
   }
}

Note that outstream files have very little chance of failing...