Re: recursively building a hash of arrays of scalars

[choroba]

my %testHash;
my $hashRef = \%testHash;
my $destref = \$hashRef->{'a'}; # Just a reference to the value, no de
+reference involved.
$$destref = "value";            # Assign to the dereference.
print $hashRef->{'a'} . "\n";

[aral]

Thank you very much! I am glad I reduced my initial problem to a minimalistic version. With your (all posters) responses, and a little bit of experimenting, I solved my problem and can now create any combination of hash, array and scalars solely working with references:

#!/usr/bin/perl

use strict;
use warnings;
use Data::Dump;

my %testHash;                # initialize hash
my $hashRef = \%testHash;    # get a reference to that

$hashRef->{'a'} = "Hello World!";

my $destref = \$hashRef->{'b'}; # Just a reference to the value, no de
+reference involved.

# initialize hash element as empty array
$$destref = [];                # NOTE: this line is not necessary

# assign a value to the hash element 'a'
@{$$destref}[0] = "How";            # Assign to the dereference.
@{$$destref}[1] = "are";            # Assign to the dereference.

# initialize array element as empty hash
@{$$destref}[2] = { };        # NOTE: this line *is* necessary    

@{$$destref}[2]->{'c'} = "you";            # assign values to hash
@{$$destref}[2]->{'d'} = "today?";        #


dd \%testHash;

# Output:
# {
#   a => "Hello World!",
#   b => ["How", "are", { c => "you", d => "today?" }],
# }

[choroba]

Line 22 is not necessary if you use the more common syntax:

my %testHash;
my $hashRef = \%testHash;
$hashRef->{a} = "Hello World!";
my $destref = \$hashRef->{b};
# assign a value to the hash element 'a'
$$destref->[0] = 'How';
$$destref->[1] = 'are';
$$destref->[2]{'c'} = "you";
$$destref->[2]{'d'} = "today?";