Why can I print this treating as a reference and treating it as a scalar?

Question

In the following perl snippet:

my $a1 = [ qw(rock pop musical) ];  
my $b1 = [ qw( mystery action drama )];  
my $c1 = [ qw( biography novel periodical)];  

my @a2d = (  
    $a1,  
    $b1,  
    $c1  
);  

The @a2d is an array which contain references to arrays.
My question is why the following print the same thing (musical)?:

print ${$a2d[0]}[2],"\n";  
print $a2d[0][2],"\n";  

I expected the second to print ARRAY or give an error since the elements of the array are refences

Answer 1

The $a2d[0] is an array reference. We can take this array reference and print out the 3rd entry:

my $ref = $a2d[0];
say ${ $ref }[2];
say $ref->[2];

These forms are equivalent. Now, we can do away with that intermediate variable, and get:

say ${ $a2d[0] }[2];
say $a2d[0]->[2];

If the dereference operator -> occurs between two subscripts, then it may be omitted as a shortcut:

say $a2d[0][2];

The arrow may be omitted when the left subscript is [...] or {...} and the right subscript it [...], {...} or (...).

This is also explained in perlreftut, which goes through these considerations in more depth. Reading that document should clear up many questions.

Answer 2

The dereference is implied when you tack on indexes.

$a2d[0][2]

is short for

${ $a2d[0] }[2]

aka

$a2d[0]->[2]

Rather than giving a syntax error, Perl provides a useful shortcut for a common operation.

Other examples: $aoa[$i][$j], $aoh[$i]{$k}, $hoa{$k}[$i] and $hoh{$k1}{$k2}.