Recipe 11.9 Constructing Records

11.9.1 Problem

You want to create a record data type.

11.9.2 Solution

Use a reference to an anonymous hash.

11.9.3 Discussion

Suppose you wanted to create a data type that contained various data fields. 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:

# Go through all records
while (($name, $record) = each %byname) {
    printf "%s is employee number %d\n", $name, $record->{EMPNO};
}

What about looking employees up by employee number? Just build and use another data structure, an array of hashes called @employees. If your employee numbers aren't consecutive (for instance, they jump from 1 to 159997) an array would be a bad choice. Instead, you should use a hash mapping employee number to record. For consecutive employee numbers, use an array:

# store record
$employees[ $record->{EMPNO} ] = $record;

# lookup by id
if ($rp = $employee[132]) {
    printf "employee number 132 is %s\n", $rp->{NAME};
}

With a data structure like this, updating a record in one place effectively updates it everywhere. For example, this gives Jason a 3.5% raise:

$byname{"Jason"}->{SALARY} *= 1.035;

This change is reflected in all views of these records. Remember that $byname{"Jason"} and $employees[132] both refer to the same record because the references they contain refer to the same anonymous hash.

How would you select all records matching a particular criterion? This is what grep is for. Here's how to get everyone with "peon" in their titles or all 27-year-olds:

@peons   = grep { $_->{TITLE} =~ /peon/i } @employees;
@tsevens = grep { $_->{AGE}   =  = 27 }      @employees;

Each element of @peons and @tsevens is itself a reference to a record, making them arrays of hashes, like @employees.

Here's how to print all records sorted in a particular order, say by age:

# Go through all records
foreach $rp (sort { $a->{AGE} <=> $b->{AGE} } values %byname) {
    printf "%s is age %d.\n", $rp->{NAME}, $rp->{AGE};
    # or with a hash slice on the reference
    printf "%s is employee number %d.\n", @$rp{"NAME","EMPNO"};
}

Rather than take time to sort them by age, you could create another view of these records, @byage. Each element in this array, $byage[27] for instance, would be an array of all records with that age. In effect, this is an array of arrays of hashes. Build it this way:

# use @byage, an array of arrays of records
push @{ $byage[ $record->{AGE} ] }, $record;

Then you could find them all this way:

for ($age = 0; $age <= $#byage; $age++) {
    next unless $byage[$age];
    print "Age $age: ";
    foreach $rp (@{$byage[$age]}) {
        print $rp->{NAME}, " ";
    }
    print "\n";
}

A similar approach is to use map to avoid the foreach loop:

for ($age = 0; $age <= $#byage; $age++) {
    next unless $byage[$age];
    printf "Age %d: %s\n", $age,
        join(", ", map {$_->{NAME}} @{$byage[$age]});

}

11.9.4 See Also

Recipe 4.14; Recipe 11.3