Section 6.5: Computer arithmetic (Frame 5)                     [prev][home][     ]

What if we look at these as unsigned numbers?

  1 0 1 1 0 1 1 1     183
+ 1 1 1 0 1 1 1 1     239
-----------------   -----
  1 0 1 0 0 1 1 0     422

But 10100110 is only 166, not 422, so another form of overflow occurred. This has to do with throwing away that final carry out. If we had kept it, we would have had 110100110, which is 422.