How is the bit length of a result of a bitwise operation for negative numbers calculated?

Question

How is the number of value bits in the result determined in the following operations:

5 = -3 & 5 # (1)11 & (0)101 -> (0)101
-3 = -3 | 5 # (1)11 & (0)101 -> (1)11
-8 = -3 ^ 5 # (1)11 ^ (0)101 -> (1)1000

?

So it seems that bitwise AND results in the biggest number of value bits between two operands and bitwise OR results in the smallest number of value bits between two operands. The bitwise XOR, however, totally doesn’t make sense. How did the 2nd lowest bit got set to 0? Why is the number of value bits bigger than the number of value bits in either operand?

Asked By: Daniil Lantukhov

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Source

Answer 1

negative number are stored in a form called two’s complement

in 4 bit system for 3 .. the representation will be 1101 and 5 is 0101

now your calculations should make sense

or

1101 
0101 
----
1101 -> -3


and

1101
0101
----
0101 -> 5


xor

1101
0101 
-----
1000  -> -8 ( in 2s complement form .. it is negative 8 not positive 8)

Answered By: ashish singh

How is the bit length of a result of a bitwise operation for negative numbers calculated?

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