Perl Cookbook

Constructing Records

Problem

You want to create a record data type.

Solution

Use a reference to an anonymous hash.

Discussion

Suppose you wanted to create a data type that contained various data fields, akin to a C struct or a Pascal RECORD. The easiest way is to use an anonymous hash. For example, here’s how to initialize and use that record:

$record = {
    NAME   => "Jason",
    EMPNO  => 132,
    TITLE  => "deputy peon",
    AGE    => 23,
    SALARY => 37_000,
    PALS   => [ "Norbert", "Rhys", "Phineas"],
};

printf "I am %s, and my pals are %s.\n",
    $record->{NAME},
    join(", ", @{$record->{PALS}});

Just having one of these records isn’t much fun— you’d like to build larger structures. For example, you might want to create a %ByName hash that you could initialize and use this way:

# store record
$byname{ $record->{NAME} } = $record;

# later on, look up by name
if ($rp = $byname{"Aron"}) {        # false if missing
    printf "Aron is employee %d.\n", $rp->{EMPNO};
}

# give jason a new pal
push @{$byname{"Jason"}->{PALS}}, "Theodore";
printf "Jason now has %d pals\n", scalar @{$byname{"Jason"}->{PALS}};

That makes %byname a hash of hashes, because its values are hash references. Looking up employees by name would be easy using such a structure. If we find a value in the hash, we store a reference to the record in a temporary variable, $rp, which we then use to get any field we want.

We can use our existing hash tools to manipulate %byname. For instance, we could use the each iterator to loop through it in an arbitrary order: