I still miss the reason of WHY although i'am redirecting all error to the new standard output which after the above code is the computer screen, i still i get both my print statememnt printed and another error as well. Why does die print 2 errors in this case?

I think you're still missing what die really does. To quote from the docs: "Outside an "eval", prints the value of LIST to "STDERR" and exits with the current value of $! (errno)." Note that the emphasis is mine: then the document goes at some length into explaining that things are not really that simple. But the point is that die exits with a value that indicates an error, independently of what gets printed to where. These are orthogonal consequences. See the following example script:

#!/usr/bin/perl

use strict;
use warnings;

if (@ARGV) {
    my $msg="This is printed to STDOUT\n";
    {
        _die => sub () {
            local *STDERR=*STDOUT;
            die $msg;
        },
        _print => sub () {
            print $msg;
            exit 0;
        }
    }->{+shift}->();
}

sub chk {
    my $cmd=shift;
    my $ret=system $0 => $cmd;
    warn "Failure to run <$cmd>: $!\n"
      and return if $ret == -1;
    warn "<$cmd> died with status: ", $? >> 8, "\n"
      unless $ret == 0;
}

chk $_ for qw/_print _die/;

__END__
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It gives me:

This is printed to STDOUT
This is printed to STDOUT
<_die> died with status: 255
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