Okay, I am fairly certain I am missing something really, really fundamental here, and my brain just won't lock on to the pattern.

I humbly ask my fellow Monks to find the proper clue × four with which to soundly smack me upside the head:

#/user/bin/perl
use strict;
my $intval = 66;
my $pckval = pack 'c', $intval;
my $pckhex = "0x" . (unpack 'H16', "$pckval");
my $stoval = "$pckval" ^ 0x80;
my $stohex = "0x" . (unpack 'H16', "$stoval");

[download]

Adding old-fashioned brute-force debugging to see what's going on:

print "\$intval = 66;  # \$intval = [$intval]  -  Expecting [66]\n";
print "\$pckval = pack 'c', $intval;  # \$pckval = [$pckval]  -  Expec
+ting [B]\n";
print "\$pckhex = \"0x\" . (unpack 'H16', \"$pckval\");  # \$pckhex = 
+[$pckhex]  -  Expecting [0x42]\n";
print "\$stoval = \"$pckval\" ^ 0x80;  # \$stoval = [$stoval]  -  Expe
+cting [194]\n";
print "\$stohex = \"0x\" . (unpack 'H16', \"$stoval\");  # \$stohex = 
+[$stohex]  -  Expecting [0xc2]\n";

[download]

This produces results other than what I'm expecting, as shown:

W:\Steve\PerlMonks>perl tpm.pl
$intval = 66;  # $intval = [66]  -  Expecting [66]
$pckval = pack 'c', 66;  # $pckval = [B]  -  Expecting [B]
$pckhex = "0x" . (unpack 'H16', "B");  # $pckhex = [0x42]  -  Expectin
+g [0x42]
$stoval = "B" ^ 0x80;  # $stoval = [128]  -  Expecting [194]
$stohex = "0x" . (unpack 'H16', "128");  # $stohex = [0x313238]  -  Ex
+pecting [0xc2]

W:\Steve\PerlMonks>

[download]

My dear fellow Monks, where have I gone astray?

In reply to Flipping the Sign Bit in pack()'ed Value by marinersk

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