how can i check if a char variable is empty?
I mean that check with like empty() method with strings, if the string doesn't contain any character stringVar.empty() will result true. How can i check if a char variable doesn't contain a character?
For example, I've a code like this:
// all libraries are included before and code is simplified because it is very long
std::fstream file;
char mychar;
file.open("userselectedfile.txt", std::fstream::in);
if (file.is_open() == true) {
   while (file.eof() != true) {
      // check if mychar is initialized yet (first time that the while execute)
      if (mychar == '') {
         mychar = file.get();
         // do something special beacuse mychar wasn't initialized
         // do something with other files
      } else {
         mychar = file.get();
         // do something else with other files
      }
   }
}
file.close();
This code isn't correct and I don't know how to solve in a nice way, i found a little thing that permise me to bypass the problem but it isn't perfect. For the moment I'm using:
std::fstream file;
char mychar;
file.open("userselectedfile.txt", std::fstream::in);
if (file.is_open() == true) {
   for (int i = 0; file.eof() != true; i++) {
      if (i = 0) {
         mychar = file.get();
      } else {
         mychar = file.get();
      }
   }
}
file.close();
Is this the only way to check if mychar isn't initialized yet? If there is a possibility to check if mychar isn't initialized yet different from the one above, which functions have I to use?
Update 1
Software aim: In this specific case the program I'm building is aimed to delete every comment in a coding source code file that the user (me cause is for me) submit, so i can't use a special character because it can be present on the file, use \0 is a nice idea but i hope that there are others. When I write //do something with ... in my program i continue to read characters till i find comments and I ignore them while creating the new file without them.
BOM: No, files i tested didn't throw me errors apart that the algorithm is a little bugged, but not the part i asked to you.
 
     
     
     
     
     
     
    