Question

How do I read in a variable number of characters? The user can input a positive or negative number that is too big to be stored in an integer. I am then checking to make sure the char is a number and storing it in an array of ints (although that will probably be changed to a short since I only need to be able to store 0-9).

istream& operator>>(istream &in, LargeInt &l)
{
  char x;
  do
  {
      cin >> x;
      int v = (int)x;

      switch( v )
      {
          case 48: v = 0;
          case 49: v = 1;
          case 50: v = 2;
          case 51: v = 3;
          case 52: v = 4;
          case 53: v = 5;
          case 54: v = 6;
          case 55: v = 7;
          case 56: v = 8;
          case 57: v = 9;
          default: v=10 /* END INPUT */;
      }

      l.largeInt[ l.usedLength ] = v;
      l.usedLength++;
      //need to check array length and make bigger if needed
  }
  while( (x == 45) || ((x > 47) && (x < 57)) );
}

Answer 1

If you wish to read digits, you need to do it a character at a time. E.g.

char ch;
while (std::cin.get(ch) && ch >= '0' && ch <= '9') {
    // You have a digit to process (maybe you want to push_back it into a vector)
}

Notice that you need to use ch - '0' to get the value of the digit because ch contains the character code rather than the value. In ASCII this means that '0' is in fact 48, '1' is 49 and so on (and ‘A’ is 65). The values may be different for different character encodings but the digits are guaranteed by the standard to be sequential, so that subtracting zero works.