Here's what I ended up with:

#!/usr/bin/perl -w
use strict;

my @list = qw(1 Mike testing 2 Richard testing2 3 David Testing3);
my @records;
my $r;

my $c = 1;
foreach my $f (@list) { 
    if ($c == 1) { $r->{id}      = $f }
    if ($c == 2) { $r->{name}    = $f }
    if ($c == 3) {
    $r->{subject} = $f;
    print "Pushing Id: $r->{id}\t Name: $r->{name}\t Subject: $r->{sub
+ject}\n"; 
# print "Anonymous hash: $r\n";
    push(@records, $r);
    $c = 1;
    $r = {};
    next;
    }
    $c++;
}

foreach (@records) {
    print "ID: $_->{id}\t";
    print "Name: $_->{name}\t";
    print "Subject: $_->{subject}\n";
}
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If you comment out the line where I assign $r to a new anonymous hash and uncomment the line where I print out the value of $r, you'll notice that it refers to the same hash. Since $r and $array[0], $array1, and $array2 are all references to the same hash, when you replace a value with one reference, you're replacing it in the one hash.

For further proof, move the right curly bracket from beneath the $c++; line to the end of the file. That will loop through @records each time and show you the change in the one hash. Solution? Make a new hash each time through the loop.

(Damian Conway describes references as "fingers pointing at the moon". You're copying fingers, but not making new moons.)