LESSON 01 – BITWISE OPERATIONS

LESSON WORK SOLUTIONS

QUESTION 1A

0011 1101

AND 0110 1100

0010 1100

QUESTION 1B

0010 1000

OR 1011 1001

1011 1001

QUESTION 1C

1001

XOR 0010

1011

QUESTION 1D

1011

NOR 1001

0100

QUESTION 1E

0101

NAND 0111

1010

QUESTION 1F

0100

XNOR 1110

0101

QUESTION 1G

NOT 1001 1011

0110 0100

QUESTION #2A

1001 1101

AND 0000 0000

0000 0000

Pattern: The value of byte A doesn’t matter as the resulting byte will always be 0s.

QUESTION #2B

1001 1101

AND 1111 1111

1001 1101

Pattern: The resulting byte is copy of byte A.

QUESTION #2C

1001 1101

OR 0000 0000

1001 1101

Pattern: The resulting byte is a copy of byte A.

QUESTION #2D

1001 1101

OR 1111 1111

1111 1111

Pattern: The value of byte A doesn’t matter as the resulting byte will always be 1s.

QUESTION #2E

1001 1101

XOR 0000 0000

1001 1101

Pattern: The resulting byte is a copy of byte A.

QUESTION #2E

1001 1101

XOR 1111 1111

0110 0010

Pattern: The resulting byte is the inversion (or 1’s complement) of byte A.

QUESTION #3A

STEP 1

We convert 17 to binary.

0001 0001

STEP 2

We now shift by 2 to the right. The shifted in bits are zeros.

0000 0100

STEP 3

We now convert back to decimal.

4

Therefore, 17 >> 2 gives 4.

VERIFICATION

Since we shifted by 2 to the right, that is the same as dividing by 2² which is 4. And, in integer division, 17 divided by 4 gives us 4. So our answer is correct!

QUESTION #3B

STEP 1

We convert 91 to binary.

0101 1011

STEP 2

We now shift by 3 to the left. The shifted in bits are zeros.

1101 1000

Note: We did shift a 1 out. So there is overflow.

STEP 3

We now convert back to decimal.

128+64+16+8=216

VERIFICATION

When we shift left by 3, that is the same as multiplying by 2³ which is 8. Unfortunately, because of the overflow, this is multiplication doesn’t match our answer.

If you really want to check your answer, you can remove the 1-bits that are shifted out in our original number. In this case, it is the 64bit that is shifted out. So, we can pretend that our original number is 91 – 64 which gives 27. Now, if we multiply 27 by 8, we get our answer of 216.

QUESTION #4A

STEP 1

Convert 73 to binary.

0100 1001

STEP 2
To get the value of -73, we first take the 1’s complement.

1011 0110

STEP 3

We now take the 2’s complement.

1011 0111

STEP 4

We now shift by 2 to the right. Since it is an arithmetic shift, we maintain the value of the leftmost bit. So, two 1s are shifted in on the left side.

1110 1101

STEP 5

We now want to go back to decimal. We must go from 2’s complement to 1’s complement by subtracting 1.

1110 1100

STEP 6

We now reverse 1’s complement.

0001 0011

STEP 7

We now convert to decimal

16+2+1=19

Therefore, the -73 >> 2 gives -19.

VERIFICATION

Unfortunately, because 2’s complement represents negative numbers by storing them off by 1, the division rule doesn’t fully work.

In this case, -73 divided by 2² which is 4 gives -18 even if the correct answer is -19.

QUESTION #4B

STEP 1

We convert 21 to binary.

0001 0101

STEP 2

We take the 1’s complement.

1110 1010

STEP 3

We take the 2’s complement.

1110 1011

STEP 4

We apply the left shift by 1.

1101 0110

STEP 5

We go from 2’s complement to 1’s complement by removing 1.

1101 0101

STEP 6

We reverse 1’s complement.

0010 1010

STEP 7

We convert to decimal.

32+8+2=42

Therefore, -21 << 1 gives -42.

VERIFICATION

Shifting to the left by 1 is equivalent as multiplying by 2. And, since -21 times 2 gives -42, we know we have the correct answer.

QUESTION #5A

1101 0110

QUESTION #5B

STEP 1

We convert 147 to binary.

1001 0011

STEP 2

We apply the circular shift to the left by three.

1001 1100

STEP 3

We convert back to decimal

128+16+8+4 = 156

The answer of circular shifting 147 << 3 is 156.

VERIFICATION

There is no convenient verification option here.

If checking your answer was very important, you could take your original number and separate it into the bits that will move up in the byte and the bits that will wrap around (or move down).

1001 0011

Green – the bits that will wrap around

Orange – the bits that will move up

The value of the green bits is 128. The value of the orange bits is 16+2+1=19.

Since the orange bits are moving to the left by three, that is like multiplying them by 2³ which is 8. So the orange bits should become 19 x 8 = 152.

Since the green bits are moving down by 5 spots, it is like dividing them by 2⁵ which is 32. So the green bits should become 128 divided by 32 = 4.

In total, the orange and green bits should total to 152 + 4 = 156 which is our answer.

QUESTION #6

STEP 1

We want to get the binary value of -122 in Java. We must recall that numbers are stored in 32 bit sequences.

We start by converting 122 to binary.

0000 0000 0000 0000 0000 0000 0111 1010

STEP 2

We take 1’s complement.

1111 1111 1111 1111 1111 1111 1000 0101

STEP 3

We take 2’s complement

1111 1111 1111 1111 1111 1111 1000 0110

STEP 4

We now do a logical shift to the right by 2. Remember that in logical shifts, the shifted-in bits are always zero.

0011 1111 1111 1111 1111 1111 1110 0001

STEP 5

Interestingly, Java sees this number as a positive number now. So can immediately convert this to decimal.

This is a little painful. I initially listed out all the values that I have to add up.

2⁰+2⁵+2⁶+2⁷+2⁸+2⁹+2¹⁰+2¹¹+2¹²+2¹³+2¹⁴+2¹⁵+2¹⁶+2¹⁷+2¹⁸+2¹⁹+2²⁰+2²¹+2²²+2²³+2²⁴+2²⁵+2²⁶+2²⁷+2²⁸+2²⁹

Not wanting to punch this into a calculator, I reformatted and entered the following directly into Google:

2^0+2^5+2^6+2^7+2^8+2^9+2^10+2^11+2^12+2^13+2^14+2^15+2^16+2^17+2^18+2^19+2^20+

2^21+2^22+2^23+2^24+2^25+2^26+2^27+2^28+2^29

And the answer that Google spit out was

1073741793 (which is the correct answer!)