bit-wise manipulation

Question 1

x^0s ^ = XOR

Answer 1

x XOR with 0s leave 0s equal to zero and ones equal to one

Question 2

x^1s

Answer 2

NOT X

Question 3

x^x

Answer 3

0s

Question 4

x & 0s

Answer 4

0s

Question 5

x & 1s

Answer 5

X

Question 6

x & x

Answer 6

x

Question 7

x | 0s

Answer 7

x

Question 8

x | 1s

Answer 8

1s

Question 9

x | x

Answer 9

x

Question 10

What is the difference between the arithmetic shift &laquo_space;and the logic shift «

Answer 10

they are similar, they shift to the left or right the bits. The arithmetic one though keeps the same most significant digits, the logic one substitute that with zeros.

Question 11

Get Bit i operation?

Answer 11

create a mask 1«i></i> then apply AND with number and see if the results is the zero or not

Question 12

clear bits

Answer 12

mask of 11110111 with not i«i></i> then you apply an end

Question 13

clever way to create a mask 000011111

Answer 13

generates a mask 000010000 with 1«i></i> then substract one

Question 14

set a bit equal to one

Answer 14

simply use the mask 00010000 with i«i></i>

Question 15

set a bit to some input value

Answer 15

You first clear by doing a non(i«i></i>) so you and this with the input then if v is your input value you or the results with v«i></i>

Question 16

How do we represent a negative number in bynary?

Answer 16

With the 2 complements. i.e we now start with 1 the sign bit. Then we use the two complement. i.e the number you have to add to the input to get to 2^bits where bits is the nu,ber of bits. You can get that by starting with the complement (0 where number is 1 and viceversa) and then add 1 to that.

Question 17

Check if number is even?

Answer 17

(num & 1) ==0

Question 18

check if number power of 2?

Answer 18

(num & num -1) == 0

Question 19

How do you check if bit array is composed by only one 1 and all zeros?

Answer 19

You can do (num - 1) and (num) ==0

0010000 –> 0001111 so they do not overlap

Question 20

How do you set the rightmost 1 bit to zero?

Answer 20

By doing x & (x-1)