Thread: Bit setting

<< Split from this 6 year old thread - Bitwise setbits function (knr 2-6) >>

Please let some one explain me how come:
n = 8
p = 16
are equally shifting the bits for 00???

But seems it is correct. I am looking forward for the answer as I really need further explanation about this.

Thanks!

Originally Posted by olbas

ok this drove me freakin insane, so i hope i can understand and implement this PITA.

x ^ (~(~0 << n) << p)
x = 1234 1234
n = 8
p = 16
~0 = FFFF FFFF

Step 1: (~0 << n) = FFFF FF00
Step 2: ~ = 0000 00FF
Step 3: << p = 0000 FF00

Step 4: x ^:
1234 1234
0000 FF00
---------------
1234 ED34

Pleaaaase tell me I got that right.

The solution included (~(~0U << n), the U represents an unsigned int type, it's not neccessary though is it?

What I am looking for is actually instead with hexadecimal numbers someone to explain with simple binary numbers for example?

Thanks!