I'm diving into bitwise operations and have used -band and -bor some, but I'm a bit lost on the others. I'm trying to find a way to set the 512 bit to 0, no matter what its current state is. I thought that maybe -bxor would do the trick, but my tests showed otherwise. Do I really need to check the bit first and then change it if needed? Here's a quick example of what I'm doing:
$number = 511
if ($number -band 512) {
$number = $number -bxor 512
}
$number
5 Answers
To set the 512 bit to 0 while keeping all the other bits intact, you can use an AND operation with the negation of 512. Here’s how you can do it:
$numberOne = 511
$numberTwo = 513
$numberOne -band (-bnot 512) # This will return 511
$numberTwo -band (-bnot 512) # This will return 1
Exactly what I needed to understand! Thanks for breaking it down.
I'm curious, but is there a case where you can't just set the number to 0 directly?
Actually, your initial approach would fail since `$number -band 512` checks if it’s already 0. If it’s not, the clause won’t run.
Unfortunately, you'll need to use an AND mask to set your specific bit to 0. For instance:
$mask = -bnot (1 -shl 9) # which is -bnot 512
$numberWith512bitZero = $number -band $mask
You can also just set it directly to 0, or alternatively, you could use `$number = $number -band 0` to achieve a similar result.
Just to clarify, the binary representation of 512 is `00000010 00000000`, and for 513 it's `00000010 00000001`. When you do -bnot 512, you effectively flip all the bits, resulting in `11111101 11111111`. This allows you to AND it with another number to keep only the desired bits.