[1,2,3] is not an array or a list. It is a reference to an anonymous array.
It is not 3 values; it�s 1 value, which happens to point to a list of size
3. If you assign that to an array via something like @a = [1,2,3], I would
expect at least a warning and possibly a compile-time error.
If it does work, it probably gets translated into @a = ([1,2,3]), which
creates an array of size 1 whose first (and only) element is a reference to
an array of size 3. That would give this result:
[EMAIL PROTECTED] == 1;
@a[0] == [1,2,3];
@a[1] == undef;
@a[0][0] == 1;
@a[0][1] == 2;
@a[0][2] == 3;
I�m not sure about +(@a[0]), but I�m guessing it would == 3.
On 2005-05-25 04:47, "TSa (Thomas Sandla�)" <[EMAIL PROTECTED]>
wrote:
> Juerd wrote:
>> > An array in scalar context evaluates to a reference to itself.
>> >
>> > A hash in scalar context evaluates to a reference to itself.
>> >
>> > An array in list context evaluates to a list of its elements.
>> >
>> > A hash in list context evaluates to a list of its elements (as pairs).
>> >
>> > Array context is a scalar context.
>
> I have understand what you mean and how you---and other p6l'er---
> derive [EMAIL PROTECTED] == 1 from @a = [1,2,3]. But allow me to regard this
> as slightly inconsistent, asymmetric or some such.
>
> Isn't hash context missing in the list above? How does
>
> %a = ( a => 1, b => 2, c => 3 ) # @a = (1,2,3)
>
> compare with
>
> %b = { a => 1, b => 2, c => 3 } # @b = [1,2,3]
>
> Does that mean
>
> 3 == +%a == +%b
> == +{ a => 1, b => 2, c => c }
> == +( a => 1, b => 2, c => c )
>
> holds and the access of the hash works as expected:
>
> %a<a> == 1 == %b<a> # and @a[0] == 1, but @b[0][0] == 1
>
> What would actually be the equivalent syntax to @b?
> Is it %b<><a> or %%b<a> or even (*%b)<a>?
> It will hardly be %b{undef}<a>, though.
