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4046

answers:

3

Hi,

how do I bind a std::ostream to either std::cout or to an std::ofstream object, depending on a certain program condition? Although this invalid for many reasons, I would like to achieve something that is semantically equivalent to the following:

std::ostream out = condition ? &std::cout : std::ofstream(filename);

I've seen some examples that are not exception-safe, such as one from http://www2.roguewave.com/support/docs/sourcepro/edition9/html/stdlibug/34-2.html:

int main(int argc, char *argv[])
{
  std::ostream* fp;                                           //1
  if (argc > 1)
     fp = new std::ofstream(argv[1]);                         //2
  else
     fp = &std::cout                                          //3

  *fp << "Hello world!" << std::endl;                         //4
  if (fp!=&std::cout) 
     delete fp;
}

Does anyone know a better, exception-safe solution?

+4  A: 

This is exception-safe:

void process(std::ostream &os);

int main(int argc, char *argv[]) {
    std::ostream* fp = &cout;
    std::ofstream fout;
    if (argc > 1) {
        fout.open(argv[1]);
        fp = &fout;
    }
    process(*fp);
}


Edit: Herb Sutter has addressed this in the article Switching Streams (Guru of the Week).

Tom
This doesn't seem any more exception-safe than the original code.
Brian
Yes it is. The original code has a memory leak if an exception is thrown while processing (*fp << "Hello World" << std::endl).
Tom
+6  A: 
std::streambuf * buf;
std::ofstream of;

if(!condition) {
    of.open("file.txt");
    buf = of.rdbuf();
} else {
    buf = std::cout.rdbuf();
}

std::ostream out(buf);

That associates the underlying streambuf of either cout or the output file stream to out. After that you can write to "out" and it will end up in the right destination. If you just want that everything going to std::cout goes into a file, you can aswell do

std::ofstream file("file.txt");
std::streambuf * old = std::cout.rdbuf(file.rdbuf());
// do here output to std::cout
std::cout.rdbuf(old); // restore

This second method has the drawback that it's not exception safe. You possibly want to write a class that does this using RAII:

struct opiped {
    opiped(std::streambuf * buf, std::ostream & os)
    :os(os), old_buf(os.rdbuf(buf)) { }
    ~opiped() { os.rdbuf(old_buf); }

    std::ostream& os;
    std::streambuf * old_buf;
};

int main() {
    // or: std::filebuf of; 
    //     of.open("file.txt", std::ios_base::out);
    std::ofstream of("file.txt");
    {
        // or: opiped raii(&of, std::cout);
        opiped raii(of.rdbuf(), std::cout);
        std::cout << "going into file" << std::endl;
    }
    std::cout << "going on screen" << std::endl;
}

Now, whatever happens, std::cout is in clean state.

Johannes Schaub - litb
Ooh, I like std::streambuf* better than std::ostream*
Tom
but if something throws, you will be left with a dangling pointer in cout of your streambuf* of the now destroyed ofstream. so beware, better write a RAII method.
Johannes Schaub - litb
I don't like hijacking std::cout. It means that cout and printf are no longer equivalent, which is something that I imagine a lot of developers take for granted.
Tom
A: 

Thanks to both of you, the examples are very helpful and solved exactly my issue.

Matthias Vallentin