从 std::cout 或 std::ofstream(file) 获取 std::ostream

Question

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?

Answer 1

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.

Answer 2

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) .

Answer 3

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

Answer 4

Referencing from this post .

You can apply a similar approach.

struct noop {
    void operator()(...) const {}
};

std::shared_ptr<std::ostream> of;
if (condition) {
    of.reset(new std::ofstream(filename, std::ofstream::out));
} else {
    of.reset(&std::cout, noop());
}

Answer 5

作为 C++ 的新手，我不知道这是否是异常安全的，但我通常这样做：

std::ostream& output = (condition)?*(new std::ofstream(filename)):std::cout;

Answer 6

The following simple code works for me:

int main(int argc, char const *argv[]){   

    std::ofstream outF;
    if (argc > 1)
    {
        outF = std::ofstream(argv[1], std::ofstream::out); 
    }

    std::ostream& os = (argc > 1)? outF : std::cout;
}