Invert bit

You are given integers 0 <= $m <= 255 and 1 <= $n <= 8. Write a script to invert bit $n from the end of the binary representation of $m and print the decimal representation of the new binary number.

Examples

Input: $m = 12, $n = 3
Output: 8

Binary representation of $m = 00001100
Invert 3rd bit from the end = 00001000
Decimal equivalent of 00001000 = 8

Input $m = 18, $n = 4
Output: 26

Binary representation of $m = 00010010
Invert 4th bit from the end = 00011010
Decimal equivalent of 00011010 = 26

Analysis

To do this I first create $mask, which is the number with only bit $n set, ie 2 ** ($n - 1).

Then if $m & $mask is greater than zero, bit $n is 1 and the desired result is $m - $mask. Conversely, if the bit is zero, the result is $m + $mask;

I am sorry that the 'Try it' feature is currently working very slowly or not at all owing to some issue with my web hosting provider.

Try it 

#!/usr/bin/perl

# Blog: http://ccgi.campbellsmiths.force9.co.uk/challenge

use v5.26;    # The Weekly Challenge - 2021-07-12
use utf8;     # Week 121 - task 1 - &&task_name
use warnings; # Peter Campbell Smith
binmode STDOUT, ':utf8';
use Encode;

invert_bit(12, 3);
invert_bit(18, 4);
invert_bit(237, 7);

sub invert_bit {
    
    my ($m, $n, $mx, $mask, $bit);
    
    # initialise
    ($m, $n) = @_;
    
    # find mask and check whether set in $m
    $mask = 2 ** ($n - 1);
    $bit = $m & $mask;

    # if bit is set in $m subtract $mask, else add it
    $mx = $bit ? $m - $mask : $m + $mask;
    
    say qq[\nInput: \$m = ] . sprintf('%3d = %08b', $m, $m) . qq[, \$n = $n];
    say qq[Output:     ] . sprintf('%3d = %08b', $mx, $mx);
}

8 lines of code
Completed after the closing date and not submitted to GitHub

Output from script

Input: $m =  12 = 00001100, $n = 3
Output:       8 = 00001000

Input: $m =  18 = 00010010, $n = 4
Output:      26 = 00011010

Input: $m = 237 = 11101101, $n = 7
Output:     173 = 10101101