Question

If there is argv[1] I want to put some data to new ofstream(argv[1]) i.e file with name argv[1]. But if there are no such argument I want to use cout instead.

I’ve tried

std::ostream& output = argc >= 1 ? std::fstream(argv[0]) : std::cout;

But it even doesn’t compile because of deleted constructor.

Answer 1

You can create an fstream instance and delay opening it until necessary.

std::fstream file;
if (argc > 1)
    file.open(argv[1]);

std::ostream& output = argc > 1 ? file : std::cout;