Question

For problems 2-4 assume all integers are in binary and in two's complement

notation. Remember to indicate overflow if necessary.

2) 0110 1010 + 1001 1110 = ?

3) 1001 1111 + 1001 0001 = ?

4) 1000 1111 + 0001 0000 = ?

Answer 1

1)
   0110 1010
   1001 1110
-------------
  10000 1000
-------------
There was an overflow of 1
and gives a result of 0000 1000
in decimal it is 8
Answer: 8 with overflow

2)
   1001 1111
   1001 0001
-------------
  10011 0000
-------------
There was an overflow of 1
and gives a result of 0011 0000
0011 0000
=>  0x2^7 + 0x2^6 + 1x2^5 + 1x2^4 + 0x2^3 + 0x2^2 + 0x2^1 + 0x2^0
=>  0x128 + 0x64 + 1x32 + 1x16 + 0x8 + 0x4 + 0x2 + 0x1
=>  0 + 0 + 32 + 16 + 0 + 0 + 0 + 0
=>  48
Answer: 48 with overflow

3)
   1000 1111
   0001 0000
------------
   1001 1111
------------
There is no overflow.
This is different, because it has 1 as left most bit. so, this number is negative.
so, to convert this to decimal, follow these steps below..
--  Flip all the bits(0's to 1, 1's to 0)
    10011111 is flipped to 01100000
--  Add 1 to the above result
    01100000 + 1 = 01100001
--  we can convert above result directly using normal binary to decimal conversion rules.
    01100001
    =>  0x2^7 + 1x2^6 + 1x2^5 + 0x2^4 + 0x2^3 + 0x2^2 + 0x2^1 + 1x2^0
    =>  0x128 + 1x64 + 1x32 + 0x16 + 0x8 + 0x4 + 0x2 + 1x1
    =>  0 + 64 + 32 + 0 + 0 + 0 + 0 + 1
    =>  97
Answer: -97 with out overflow