Question

I have two data structures in Perl:

An array:

my @array2 = ( "1", "2", "3");

for $elem (@array2) {
    print $elem."\n";
}

Giving me the following output:

1
2
3

And a list:

my @array = [ "1", "2", "3"];                                            

for $elem (@array) {
    print $elem."\n";
}

Giving the following output:

ARRAY(0x9c90818)

Obviously, I’d like to iterate over the elements in both cases, but why does the second solution give me only the reference to this array?

Answer 1

Lists in Perl are not data structures, they are positions in the source code, determined by the context around them. Lists are basically the transient structures that Perl uses to move data around. You interact with them with all of Perl’s syntax, but you can not work with them as a data type. The data type that is closest to a list is an array.

my @var    =    (1, 2, 3);  # parens needed for precedence, they do not create a list
   ^ an array    ^ a list

say 1, 2, 3;
    ^ a list

say @var;
    ^ a list (of one array, which will expand into 3 values before `say` is called)

When you write [1, 2, 3] what you are doing is creating a scalar array reference. That array reference is initialized with the list 1, 2, 3, and it is the same as creating a named array and taking a reference to it:

[1, 2, 3]   ~~   do {my @x = (1, 2, 3); \@x}

Since the [...] construct creates a scalar, you should hold it in a scalar:

my $array = [1, 2, 3];                                            

for my $elem (@$array) {   # lexical loop variable
    print $elem."\n";
}

Since you want to operate on the whole array, and not just the reference, you place a @ in front of the $array which dereferences the stored array reference.