# Re: Plethora of operators

```On 5/14/05, Larry Wall <[EMAIL PROTECTED]> wrote:
> On Sat, May 14, 2005 at 01:36:22PM -0500, Rod Adams wrote:
> : Larry Wall wrote:
> :
> : >On Sat, May 14, 2005 at 12:51:32PM -0500, Rod Adams wrote:
> : >
> : >: Unless, of course, there is some subtle difference between a 3-d hash
> : >: and a hash of hashes of hashes that invalidates this.
> : >
> : >No difference, I hope.  The multidimensional notation is meant
> : >to extend to HoH and AoA transparently (as well as HoA and AoH).
> : >Since a variable's dimensionality is always declared (and a container
> : >object's dimensionality generated at "new" time), we shouldn't have
> : >to worry about whether to add dimensions or autovivify a reference.
> : >Either there's storage already allocated, or we autovivify.
> : >
> : >
> : Hmm. So if I say:
> :
> :    @a = [ { a => 1, b => 2}, { a => 3, b => 4 } ];
> :
> : Can I then say:
> :
> :    \$x = @a[1;'b'];
> :
> : And get \$x = 4?
>
> Probably not, but @a{1;'b'} might.  I think what we've said before is
> that .[] allows the optmimizer to assume numeric subscripting only,
> while .{} is the more general form.```
```
So, does this mean that I can do something like:

@a = [ 1 .. 4 ];
\$x = @a{2};

and have \$x == 3? If so, is there any reason (other than clarity) to
use the @a[] notation? The @ already indicates you have an array vs.
the % which indicates hash. Is there a reason to have the subscripting
notation also be different?

I understand why it was different in P5, given that you needed to
differentiate \$x->[2] and \$x->{2} and allow the reader to know whether
\$x was an arrayref or hashref. But, that need is gone from P6. (Isn't
it?)

Rob
```