I came up this example, which could give a very intuitive answer:

use strict;
use warnings;

my $a = "abcdabcd";
{
    my $b = join('.',unpack('CCCC',$a));
    print $b, "\n";
}
{
    my $b = join('.',unpack('CCCC',substr($a,0,4)));
    print $b, "\n";
}
{
    my $b = join('.',unpack('(CCCC)*',$a));
    print $b, "\n";
}
{
    my $b = join('.',unpack('(CCCC)*',substr($a,0,4)));
    print $b, "\n";
}
[download]

The output is:

97.98.99.100
97.98.99.100
97.98.99.100.97.98.99.100
97.98.99.100
[download]

Here is the explanation:

1. The pattern in the first block says: take the first 4 bytes, and interprete each of them as unsigned char. The bold part is an implicit substr.
2. The second block uses the same pattern, but it substr $a first. Base on the explanation of the first block, now we know that what the program does is "take the first 4 bytes of the first 4 bytes", which has no difference from 'take the first 4 bytes'. So this produces the same thing as case 1.
3. Now we switch to a new pattern '(CCCC)*', which says repeatedly group every 4 consecutive bytes together. Without substr, as $a is 8 bytes long, it matches the pattern twice, or more precisely once but produces two groups. So it uses all 8 bytes.
4. Same pattern as case 3, but substr'd down to 4 bytes first, now you only have enough bytes to match the pattern once and produce one group.

Replace

    (1 x $str2), (1 x (32 - $str2))
[download]

with

    (1 x $str2) . (0 x (32 - $str2))
[download]

B32 want a string of 32 characters (preferably where each character is "0" or "1"). When you give it a string of fewer than 32 chars, then it assumes "0"s for the missing chars. Your second string is simply being ignored.

$mask = pack ('B32', (1 x $2), (1 x (32 - $2)));
$new {ISPP} {inet_aton ($1) & $mask} =                                
+                                
               $mask . pack ('B32', (1 x 32));
[download]