perl - trouble understanding array of arrays

Question

I am trying to make an array of arrays and then reference them.

I make somethign like:

sub foobar
{
 my @array;
 my $i;
 for ($i = 0; $i < 1000; $i=$i+1)
 {
  my @row;
  $row[0] = $i;
  $row[1] = foo($bar);
  push @array , [@row];
 }
 return \@array;
}

I can get to the values via:

$array->[x]->[y];

However I don't understand why the second -> is needed. $array->[x] I understand because $array is a reference. But isn't $array->[x] meant to be an array? Why doesn't this work:

my @notarray = $array->[x];

What exactally is not array filled with now? Because it certainly doesn't seem to be an array containing $i , foo($bar)

How would $array->[x]->[y] be different for a reference to an array of references to arrays?

Platinum Azure · Accepted Answer · 2011-02-10 04:25:38Z

4

The second -> isn't needed, actually.

Here's the deal: All Perl array values and hash values must be scalars. That means either a string, number, or array/hash reference (and not a plain old array or hash).

So the first -> operator dereferences the array and gets at the x'th row. In there is-- not an array, but an array reference. So in order to get to the data in there, you'd theoretically need another -> operator.

But get this. Perl is smart: It knows that after one array or hash access, if another access happens, the only way this is possible is through an array/hash reference (because your first array/hash access MUST return a scalar)! So you don't need the second arrow after all.

See perldata for more details.

answered Feb 10, 2011 at 4:25

Platinum Azure

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1 Comment

kiasectomondo Over a year ago

Thanks, for some reason I had come to believe that it was possible to have arrays of arrays, but they are really arrays of references.

Ted Hopp · Accepted Answer · 2011-02-10 04:18:11Z

2

When you

push @array, [@row];

you are pushing a reference to an array. This is necessary because of Perl's rule that arrays are flattened. So $array->[x] is a reference to the row array, not the row array itself. However, between subscripts, the arrow is optional. So $array->[x]->[y] is exactly the same as $array->[x][y] (which is exactly the same as ${$array}[x][y], etc.)

This is all explained in the Perl reference tutorial

answered Feb 10, 2011 at 4:18

Ted Hopp

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3 Comments

kiasectomondo Over a year ago

If I am pushing a reference then why is each row still different? Shouldn't they be refering to the same thing?

Platinum Azure Over a year ago

@kiasectomondo: You're creating a fresh anonymous reference every time. If you just pushed \@array each time (for the same @array) that would refer to the same thing.

Eric Strom Over a year ago

@kiasectormondo => The rows would only refer to the same thing if you had declared the @row array outside of the loop, and taken a reference to it via \@row. The [@row] syntax creates a shallow copy of the array, and returns the new reference. In your case, since you are declaring my @row inside the loop, you can just use \@row and it will work fine.

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