CODEWITHSUNDEEP
twos-complement
Dimitri
Dimitri

Reputation: 1966

Two's complement confusion with certain number of bits

Say I have 21 - 10 = 10 in decimal.

Signed method:

 10101
- 1010
 -----
 1011 = 11 (in decimal)

So far so good,

Now if I decide to use two's complement

     (0) 
      10101 (21)
    + 10110 (-10)
      -----
  (1) 01011 = (11)

Here it shows I have a carry in of zero and carry out of one. If I follow the rules of underflow and overflow, this should be considered as an underflow and should not be getting the correct answer in the end.

However, if I decide to add one more bit to work with:

     (1)
      010101 (21)
    + 110110 (-10)
      -----
  (1) 001011 = (11)

In this case, I get the correct answer and no overflow.

I cannot seem to figure out what's happening here and why is it that I'm getting the correct answer for each of these cases in two's complement.

Any clarification would help!

Upvotes: 1

Views: 557

Answers (2)

The Vermilion Wizard
The Vermilion Wizard

Reputation: 5395

With only 5 bits, 10101 is the representation of -11. You're getting the "right" answer because what you're actually calculating is -11 + -10 and getting 11, which is the incorrect answer, consistent with the rules of over/underflow.

Upvotes: 2

Mike Nakis
Mike Nakis

Reputation: 61949

In 5 bits, 10101 is not 21!

The most significant bit is set, so it is some negative number!

Upvotes: 2

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