Re: How can I set a bit to 0 ?

The XOR operator works just like the bitwise NOT operator (~) because it flips bits, but it takes two inputs. Input number 1 determines which bits will be negated in input number 2:

For example, 0000 ^ 0000 will simply leave all the bits zero.
1111 ^ 0000 will flip all the bits, so they become 1111.
1000 ^ 0000 will flip only the first bit and leave the rest unchanged. Result: 1000
1110 ^ 1101 will flip the first three bits and leave the last one unchanged. Result: 0011

The AND operator turns specific bits OFF. It takes two input numbers. The bits in input number 1 determine which bits in input number will be turned off. The ones that are zero will be turned off. Here is an example:

1001 ^ 0110 = 0000
1100 ^ 0111 = 0100
0000 ^ 1111 = 0000
1000 ^ 1111 = 1000

See, this is really simple. Here is a tiny example program:

#!/usr/bin/perl -w

use strict;
use warnings;


$a = 4;

print "\n ORIGINAL = $a";

$a &= 0xFFFB;    # 0xFFFB = 1111111111111011 binary

print "\nCLEAR BIT = $a";

$a |= 4;

print "\n  SET BIT = $a";

print("\n  XOR BIT = ", ($a ^ 0));

print("\n  XOR BIT = ", ($a ^ 0xF));

exit;
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Re^2: How can I set a bit to 0 ?
by AnomalousMonk (Archbishop) on May 28, 2022 at 04:49 UTC

Input number 1 determines which bits will be negated in input number 2 ...

Just a note of caution: This phrasing suggests that the order of the operands affects the result of the ^ (bitwise-xor) operator, but it does not. The order is irrelevant:

Win8 Strawberry 5.8.9.5 (32)  Sat 05/28/2022  0:35:22
C:\@Work\Perl\monks
>perl

use strict;
use warnings;

printf "%04b \n", 0b1100 ^ 0b0111;
printf "%04b \n", 0b0111 ^ 0b1100;

^Z
1011
1011
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|

&

Update: Slight change to phrasing of first sentence for clarity.

Give a man a fish: <%-{-{-{-<

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