MHB Bit Masking: How to Manipulate Bits in Binary Numbers

[gEOdude]

Hey Guys, can i have a hand with this question. I just want to know if what I did was right.

Bit masking has a task. They are used in order to access specific bits in a byte of data.

1) Using the 8-bit binary number “1001 1101”:
Turn off the 2 bits (masking bits to 0) on each side, and then leave the middle 4 alone.The output should be achieved should be "0001 1100”.

2) Using the 8-bit binary number “1001 1101”:
Toggle the values of the middle 4 bits (the opposite of what it currently is), and leave the 2 bits on each side untouched.The output achieved should be"10100001”. Answer
1) (1001 1101) & (0101 1110) = 0001 1100
2) (1001 1101) ^ (0111 1110) = 1010 0001

 

[Evgeny.Makarov]

1) Using the 8-bit binary number “1001 1101”:
Turn off the 2 bits (masking bits to 0) on each side, and then leave the middle 4
alone.

Toggle the values of the middle 4 bits (the opposite of what it currently is), and leave the 2 bits on each side untouched.

Why do these quotes talk about the middle 4 bits and the 2 bits on each side, that is, 6 bits in total, when the input number consists of 8 bits?

 

[gEOdude]

Im not sure, its how my teacher wrote the problem. Is there something wrong with it?

 

[I like Serena]

Bit masking has a task. They are used in order to access specific bits in a byte of data.

1) Using the 8-bit binary number “1001 1101”:
Turn off the 2 bits (masking bits to 0) on each side, and then leave the middle 4 alone.The output should be achieved should be "0001 1100”.

The mask that masks out 2 bits on each side is [m]0011 1100[/m].
Using this mask we can do:
1001 1101 & [m]0011 1100[/m] = 0001 1100

2) Using the 8-bit binary number “1001 1101”:
Toggle the values of the middle 4 bits (the opposite of what it currently is), and leave the 2 bits on each side untouched.The output achieved should be"10100001”.

Using the same mask:
1001 1101 ^ [m]0011 1100[/m] = 1010 0001

Answer
2) (1001 1101) ^ (0111 1110) = 1010 0001

I'm afraid the xor operator is not evaluated properly here. (Worried)

Why do these quotes talk about the middle 4 bits and the 2 bits on each side, that is, 6 bits in total, when the input number consists of 8 bits?

Im not sure, its how my teacher wrote the problem. Is there something wrong with it?

No.
2 bits on each side means 2 bits to the left and 2 bits to the right, for 2+4+2=8 bits in total.

 

[Evgeny.Makarov]

2 bits on each side means 2 bits to the left and 2 bits to the right, for 2+4+2=8 bits in total.

Of course. Sorry about my misunderstanding.