It's a matter of opinion, but I believe insisting on solving with modules will hinder your ability to code. This one seems simple enough. Just convert to binary before comparing input with binary-converted limits of the range :- 
my $limit1 = '101.0.0.0';
my $limit2 = '101.0.255.255';

my ($bl1,$bl2) = (cnvb($limit1),cnvb($limit2));

while (<>) {
    chomp;
    my $bin = cnvb($_);
    if ( $bin >= $bl1 and $bin <= $bl2 ) {
        print "$_\n";
    }
}

sub cnvb {
    my @mask = split /\./, shift();
    $#mask == 3 or die "bad IP address";
    my $ret = 0;
    for my $i ( 0 .. 3 ) {
        $ret *= 256;
        $ret += $mask[$i];
    }
    return $ret;
}

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One world, one people

In reply to Re: New to Perl by anonymized user 468275
in thread New to Perl by NastyPenguin

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