I'm trying to assign an array to a value in my hash as follows:
$authors->[$x]->{'books'} = @books;
$authors is an array of hashes that contains his/her first name, last name, date of birth, etc. And now I'm creating a books key where I want to assign an array of books. However, when I try to print it afterwords, it's just printing the size of the array, as if I'm doing $value = scalar @books.
What am I doing wrong?
Array elements and hash values are scalars, so when you are nesting arrays and hashes, you must use references. Just as $authors->[$x] is not a hash but a reference to a hash, you must set $authors->[$x]->{'books'} to a reference to the array.
$authors->[$x]->{'books'} = \@books; # reference the original array
$authors->[$x]->{'books'} = [@books]; # reference a copy
You would then access elements of the array using something like
$authors->[$x]->{'books'}->[0]
which can be abbreviated
$authors->[$x]{books}[0]
or access the whole array as
@{$authors->[$x]{books}}
Your original attempt
$authors->[$x]->{'books'} = @books;
is exactly equivalent to
$authors->[$x]->{'books'} = scalar @books;
because the left operand of the = operator is a hash value, which is a scalar, so the right operand is evaluated in scalar context to provide something that can be assigned there.
P.S.
On rereading this answer I realized it may be confusing to say "a hash value is a scalar" because of the possible interpretation of "hash value" as meaning "the value of a hash variable" i.e. "the whole hash". What I mean when I write "hash value" is an item that is stored in a hash as a value (as opposed to a key).
[ @books ] (you can call [ ]the anonymous array constructor) takes @books and makes a copy that can be held as a reference "anonymously". With \@books you are referring to the original array. This makes a difference if you change any of the elements - i.e.does it change the copy or the reference and the original? See perldoc perlreftut and perldoc perlref. There's many a handy guide/blog post comparing perl references and C pointers :-)While the first awnser is absolutely right, as an alternative, you could also fo this:
push @{$authors->[$x]->{'books'}}, @books;
Then $authors->[$x]->{'books'} will be an Array that contains a copy of all
the elements from @books.
This might be more "foolproof" then working with references, as
mentioned above.
$authors->[$x]->{'books'} = [@books]; # reference a copy. My bad, should have looked better at the other awnser.