8 bit unsigned number addition and subtraction overflow

Question

I have trouble understanding when overflow occurs in unsinged addition and subtraction. For example,

   1 1 1 1 0 0 0 0 
+  0 0 1 1 1 0 0 0 
  __________________
   0 0 1 0 1 0 0 0

Because of the 1 in the MSB, it results as overflow. I understand this problem, but when it gets to subtraction, I have trouble determining when it is overflow

For example,

   0 0 0 0 0 0 0 1               
-  0 0 0 0 0 0 1 1

(After applying 2's complement)

   0 0 0 0 0 0 0 1 
+  1 1 1 1 1 1 0 1

   1 1 1 1 1 1 1 0

Therefore the result should be "no overflow" because there is no 1 carry out in the end. However, the answer says "overflow". Could you please tell me why?

I think this questions should be asked on - electronics.stackexchange.com — Germa Vinsmoke
– Germa Vinsmoke, Commented Sep 23, 2018 at 18:06

MvG · Accepted Answer · 2018-09-23 13:16:08Z

1

I'd probably call this underflow not overflow.

Think about it like this. You are trying to compute x − y. Using two's complement you are implementing this as x + (2^N − y) = 2^N + (x − y). So the result will only correctly represent (x − y) if there is an overflow carry bit 2^N that fell off the left. Otherwise the result you get is the two-s complement representation of a negative subtraction result.

answered Sep 23, 2018 at 13:16

MvG

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