Thread: Bit Wise Functions

Shift right first BITS - LEFTSHIFT times. Put that into a variable. OR that variable with the original left shifted LEFTSHIFT times.


Quzah.

What you are trying to do is fairly straight-forward, if I understand the problem properly. To change a 1 into a 0 at position n, we simply take num1^(1<<n) (note: using XOR like this will change a 1 into 0, but it will also change a 0 into 1, so it doesn't guarantee that there will be a 0 at that point after you do the XOR, only that it will be different than it was before). To change a 0 into a 1 at position n we simply take num1|(1<<n) (this always will leave a 1 at position n, even if it is already a 1, since we use OR and not XOR). Here is an illustration of your specific example:

Code:

num1		10100000000000000110001100100000
		00000000000000000000000000001000 	|
		-------------------------------------
num1		10100000000000000110001100101000


num1		10100000000000000110001100101000
		00000000000000000000000000100000	^
		-------------------------------------
num1		10100000000000000110001100001000

We could have used XOR both times and gotten to the same result.

I'm not sure understand what you are wanting to do.
Are you wanting to move bit 5 to where bit 3 is, setting bit 5 to zero in the result? (count bits starting from: 0 is the rightmost one)

1. If so, first you want to get the original value with the desired bits zeroed out.
2. Next you want get the desired bit isolated and shift it along.
3. Lastly you want to combine that bit with the value from step 1.

Some code for this:

Code:

unsigned int a = input & ~0x28;       // 0x28 => 00101000 where those 1's are the bits we want cleared
unsigned int b = (input & 0x20) >> 2; // 0x20 => 00100000 where the 1 is the bit we want to isloate
output = a | b;

Or as a one-liner:

Code:

output = (input & ~0x28) | ((input & 0x20) >> 2);