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