Difference between () and [] in Perl 6

Question

Note This SO should not have the raku tag. It is too obsolete/misleading. The technical problem discussed in the question body no longer applies. The disagreement in the comments about naming/tags no longer applies. I'm leaving it for historical interest only, under the old tag only.

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I am learning Perl 6, and had trouble understanding the Perl 6 one-liner below

My Perl 6 is rakudo-star: stable 2014.04 (bottled)

This works fine. The array/list is sorted

[njia@mb-125:~] : perl6 -e 'say [2443,5,33, 90, -9, 2, 764].sort'
-9 2 5 33 90 764 2443

But this does not sort the array/list, if [].sort works why @s.sort does not?

[njia@mb-125:~] : perl6 -e 'my @s = [2443,5,33, 90, -9, 2, 764]; @s.sort.say'
2443 5 33 90 -9 2 764

Change from [] to ()

[njia@mb-125:~] : perl6 -e 'my @s = (2443,5,33,90,-9,2,764); @s.sort.say'
-9 2 5 33 90 764 2443

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NOTE the described behavior in this question has changed in the release version of perl6. See response by G. Cito below.

Answer 1

For those who may be confused by this answer, the question is about Perl 6, and none of this applies to Perl 5.

The statement

my @s = [2443, 5, 33, 90, -9, 2, 764]

creates an itemised array and assigns it to @s[0], so @s has only a single element and sorting it is pointless.

However you can say

@s[0].sort.say

which has the effect you expected

Answer 2

I'm going to go out on a limb and refer to some of CPAN's Perl6 documentation where this could be viewed as a list vs. array thing - i.e. a sequence of values versus a sequence of itemized values (see doc.perl6.org).

Certainly perl6 is different enough that it warrants its own tag but it is still perl so it's not surprising that () creates a list and [] creates an anonymous array.

> say [2443, 5, 33, 90, -9, 2, 764].WHAT
(Array)
> say (2443, 5, 33, 90, -9, 2, 764).WHAT
(List)

Since this question was first asked and answered the behavior has changed:

> my @s = [2443, 5, 33, 90, -9, 2, 764]
> @s.sort.say
(-9 2 5 33 90 764 2443)

Note that the output when sorted is a List but otherwise @s is an Array:

> @s.sort.WHAT.say
(List)
> @s.WHAT.say
(Array)