Discrimination between file and console streams

Question

How to determine weather ostream is a file or a console stream. In the following program I want to print "Hello file!" while writing to a file and "Hello console!" while writing to console. What condition should I specify at line 17?

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

class A{
public:
        A(string msg):_str(msg){}
        string str()const {return _str;};
private:
        string _str;
};

ostream & operator << (ostream & os, const A & a)
{
        if (os is ofstream) //this is line 17
                os << "Hello file! " << a.str() << endl;
        else
                os << "Hello console! " << a.str() << endl;

        return os;
}

int main()
{
        A a("message");
        ofstream ofile("test.txt");
        if (!ofile)
                cerr << "Unable to open file";
        else
                ofile << a;  // "Hello file"

        cout << a << endl; // "Hello console"
}

Answer 1

Maybe not pretty, but

std::streambuf const * coutbuf = std::cout.rdbuf();
std::streambuf const * cerrbuf = std::cerr.rdbuf();

ostream & operator << (ostream & os, const A & a)
{
        std::streambuf const * osbuf = os.rdbuf();

        if ( osbuf == coutbuf || osbuf == cerrbuf )
                os << "Hello console! " << a.str() << endl;
        else
                os << "Hello file! " << a.str() << endl;

        return os;
}

We could use &os == &std::cout, but the Standard output might be redirected to file, so I think it is better to use the streambuf object instead. (See this answer for better understanding as to how the redirection works, and why comparing streambuf solves the problem safely! )

Answer 2

You could (ab)use tellp(), which returns -1 if the stream does not have a position:

bool isConsoleStream(ostream const& stream)
{
    return stream.tellp() == -1;
}

Of course, there could be other streams that return -1 for this function, so use with caution.

Answer 3

There is no portable means. Under Unix, you can do:

if ( (&os == &std::cout && isatty( STDOUT ))
        || (&os == &std::cerr && isatty( STDERR ))
        || (&os == &std::clog && isatty( STDERR )) ) }
    //  is a terminal...
}

Under Windows, the isatty becomes _isatty, and I'm not sure that the macros exist (but I suspect that they do).

Of course, this supposes that you don't do things to confuse it in your code. Something like:

std::ostream s( std::cout.rdbuf() );

for example, or:

std::cout.rdbuf( &someFileBuf );

Or even:

std::ofstream s( "/dev/tty" );  //  (or "CONS" under Windows).

But it's about as close as you can get without the actual fd from the filebuf.

Answer 4

One is a ofstream and the other is a ostream. Just have two methods.

#include <iostream>
#include <string>
#include <fstream>

class A {
    std::string s;
public:
    A(const std::string& s) : s(s){}
    std::string str() const {return s;}
};


ostream & operator << (std::ostream & os, const A & a)
{
    return os << "console: " << a.str() << std::endl;
}

ofstream & operator << (std::ofstream & os, const A & a)
{
    return os << "file: " << a.str() << std::endl;
}

int main()
{
    A a("hello world");
    std::cout << a << endl;
}