How to redirect std::cout to file conditionally

Question

In my code I have a a function that takes as argument a boolean variable. If this variable is true then the output shall be redirected to a file. If not, then to std::cout

The code looks like this and Its been inspired by a relevant question [1]

void MyClass::PPrint(bool ToFile)
{
    
        std::streambuf *coutBufferBak = std::cout.rdbuf();
        std::ofstream   out("out.txt");
        
        if (ToFile) { std::cout.rdbuf(out.rdbuf()); }

        std::cout << "Will I be written to a file or to cout ?\n";

        if (ToFile) {std::cout.rdbuf(coutBufferBak);} // reset cout
  
}

But in the case that the ToFile flag is false the file will be generated nonetheless and it will be empty. Is there a way to do it so that the file won't be generated ? If for example I try to include on the first IF statement the std::ofstream out("out.txt"); then I will get a SegFault due to the the variable scope being limited to that if.

Why can't you simply defer the opening/creation of `out.txt` until the body of the `if (ToFile)` conditional? — G.M., Mar 15 '21 at 12:16
you need not redirect `std::cout` when you can just choose between `std::cout` and a file stream — 463035818_is_not_an_ai, Mar 15 '21 at 12:16
It looks like you should wrap the `rdbuf` logic into a class. This gives you RAII behavior; the dtor will automatically reset `cout`. You can then use a `std::optional<>` wrapper to handle the conditional part. — MSalters, Mar 15 '21 at 12:28
FWIW, I would change the function to `void MyClass::PPrint(ostream& out)`, then the function body just becomes `out << sutff_to_output`. This allows users of the function to print to any stream, and also the ability to pick a different file location. — NathanOliver, Mar 15 '21 at 12:35

Caleth · Accepted Answer · 2021-03-15T13:14:00.820

3

Don't redirect std::cout. Write your print in terms of a std::ostream & parameter, and choose an appropriate std::ostream to pass.

void MyClass::PPrintImpl(std::ostream & out)
{
    out << "Will I be written to a file or to cout ?\n";
}

// a.k.a 

std::ostream & operator <<(std::ostream & out, const MyClass &)
{
    return out << "Will I be written to a file or to cout ?\n";
}

void MyClass::PPrint(bool ToFile) {
    if (ToFile) {
        std::ofstream fout("out.txt");
        PPrintImpl(fout);
    } else {
        PPrintImpl(std::cout);
    }
}

We pass std::ostreams by reference because we the identity, not just the value of the stream object matters. We know this because they aren't copyable (the copy constructor is deleted)

edited Mar 15 '21 at 13:14

answered Mar 15 '21 at 12:35

Caleth

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Clean programming wise, this seems to me the best practice. Thank you. I've been quite struggling with ```c++``` and this seems very handy. Could you elaborate maybe on why the ```operator ``` is needed in this case if possible? And why the ```ostreams``` must be passed as refferences to the functions.Thanks again. – ex1led Mar 15 '21 at 13:00
1

@Broxigar It isn't *needed*. It's just that the conventional name of a "write to output stream" function is `operator <<` – Caleth Mar 15 '21 at 13:02

score 2 · Answer 2 · answered Mar 15 '21 at 12:16

2

You can open the file conditionally:

std::ofstream out;
if(...)
    out.open("out.txt");

answered Mar 15 '21 at 12:16

Daniel

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How to redirect std::cout to file conditionally

2 Answers2