Is there a function in C++ that creates a .bin file, or is this code missing something?

Question

I have a code that looks like this:

int main () {
  fstream file;
  file.open("test.bin", ios::out | ios::binary);
  if(!file.is_open())
  {
      return -1;
  }
  int n = 3;
  file.write(reinterpret_cast<char*>(&n), sizeof(n));
  file.close();
  return 0;
}

when I run it alone, it exits with -1, so obviously it failed to open "test.bin". However, if I save a blank notepad file as "test.bin", and run it, it works fine. What I am wondering is how I can get my C++ program to automatically generate an empty file "test.bin" if a file called "test.bin" does not already exist.

Answer 1

I'd assume you could probably just do it by opening and closing a file:

if (GetFileAttributes("test.bin") == INVALID_FILE_ATTRIBUTES)
{
    fstream file;
    file.open("test.bin", ios::out);
    file.close();
}

Answer 2

One option is to open for read/write and seek to the beginning of the file.

Then, you may read, write, or do whatever you wish.

Answer 3

Your code snippet is wrong since it's trying to write to a file that you've opened for input. If you want to write to the file, simply use ios::out instead of ios::in.

If you want to open the file for reading but create it if it does not exist, you can use:

file.open("test.bin", ios::in | ios::binary);
if(!file.is_open()) {
    file.open("test.bin", ios::out | ios::binary);
    int n = 3;
    file.write(reinterpret_cast<char*>(&n), sizeof(n));
    file.close();
    file.open("test.bin", ios::in | ios::binary);
    if(!file.is_open()) {
        return -1;
    }
}

This will initialize the file with the integer 3 as the default contents, if it doesn't already exist.

If it does exist, it will leave the contents alone. In either case, you'll have the file open at the first byte.