title: Add integers given by 16 bits in two's complement form (long integer addition) given words of only 8 bits
body: <p>I have been struggling with a problem for hours and can't seem to wrap
  my mind around this properly.</p>
  
  <p>I'm using a program that's simulates a Computer architecture with 4-bit bytes, 2-byte words, 16 general purpose 2-byte registers, and a fixed
  4-byte instruction format. Words align on 2-byte boundaries and instructions on 4-byte boundaries.
  Integers are stored in two’s complement format.</p>
  
  <p>Every byte is represented in hexadecimal form, e.g. 0xff = -1, and so on.
  So a word is seen as ff, 01, a1, etc in the RAM memory.</p>
  
  <p>I'm trying to add two four-bytes integers in two's complement notation
  (long integer addition). That is given an input of two 16-bit words X1;X0; Y1; Y0 outputs a 16-bit word Z1Z0 = X1X0 +
  Y1Y0.</p>
  
  <p>I understand that's the integers don't fit in the given range for a word, 
  and my thoughts after reading a lot on about long integer addition
  is that one should divide it into smaller parts and add "digit" for digit.</p>
  
  <p>Won't there by some problem with the eight bit with only is a signed bit in this case? I must admit that my mind is quite lost at the moment, so hope it's understandable.  </p>
  
  <p>Best Regards 
  A curious mind</p>