Panic - Y2k38 Bug!!!


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Thread: Panic - Y2k38 Bug!!!

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  1. #1
    Join Date
    Jul 2001
    Location
    Fife, Scotland
    Posts
    1,794

    Panic - Y2k38 Bug!!!

    Try pasting the following code into a standard Linux file, change permissions so it's executable and then run it:

    Code:
    #!/usr/bin/perl
    #
    # I've seen a few versions of this algorithm
    # online, I don't know who to credit. I assume
    # this code to by GPL unless proven otherwise.
    # Comments provided by William Porquet, February 2004.
    # You may need to change the line above to
    # reflect the location of your Perl binary
    # (e.g. "#!/usr/local/bin/perl").
    # Also change this file's name to '2038.pl'.
    # Don't forget to make this file +x with "chmod".
    # On Linux, you can run this from a command line like this:
    # ./2038.pl
    use POSIX;
    # Use POSIX (Portable Operating System Interface),
    # a set of standard operating system interfaces.
    $ENV{'TZ'} = "GMT";
    # Set the Time Zone to GMT (Greenwich Mean Time) for date calculations.
    for ($clock = 2147483641; $clock < 2147483651; $clock++)
    {
        print ctime($clock);
    }
    # Count up in seconds of Epoch time just before and after the critical event.
    # Print out the corresponding date in Gregorian calendar for each result.
    # Are the date and time outputs correct after the critical event second?
    We're all going to DIE!!!

    (Well, maybe in 31 years time... by which time they MIGHT have found a solution to the problem).

    Sorry to be a misery, but:

    THE WORLD'S GOING TO END!!!

    James

    PS. Your thoughts???
    Last edited by Satanic Atheist; 07-26-2007 at 02:45 PM. Reason: Messed up HTML tags
    -----------------------------
    UseLinux.net
    -----------------------------

    perl -e 'use Math::Complex;$|=1;for$r(0..24){for$c (0..79){$C=cplx(($c/20.0)-3.0,-($r/12.0)+1.0);$Z= cplx(0,0);for($i=0;($i<80)&&(abs($Z)<2.0);$i++){$Z =$Z*$Z+$C;}print$i>=80?"*":" ";}print"\n";}'

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