Re: Hash initialization works, but why?

The short answer is that @array{1,2,3} is a hash slice, and therefore an array. You can tell it's a hash slice because the indexes are inside {curlies}. You can tell it's an array because it starts with an @.

It's rather like the whole beginner array-vs-array-item confusion that so many people encounter, at least briefly: if @x is my array, surely item 3 in @x should be @x[3] ?

But of course it's not, it's $x[3] because the item itself is a scalar.

In this case, even though we're retrieving from (or in your case assigning to) a hash, we're interacting with it in a list-centric way rather than a one-thing-at-a-time way.

So @x{...} refers to %x. It has nothing to do with @x at al, and also nothing to do with typeglobs.

The trick is to understand that the kind of brackets around the index ({} vs []) determine whether it's an array or hash that you're dealing with, not the symbol at the front, which only determines how many things you're dealing with in one operation.

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Re^2: Hash initialization works, but why? by ikegami (Patriarch) on Jun 04, 2008 at 23:09 UTC
The short answer is that @array{1,2,3} is a hash slice, and therefore an array That's wrong. It a hash slice, and it produces a list. It's not an array. Very different.	[reply]
Re^3: Hash initialization works, but why? by throop (Chaplain) on Jun 04, 2008 at 23:23 UTC
I admit, I've always thought that 'array' was just Perl's particular jargon for 'list'. Or maybe "Perl's data-structure for holding a list." If I write `@foo = (1,2,3)` [download] then @foo is an array, while (1,2,3) is a list. Right? I find myself still confused on the distinction between an array and a list. Could you explain more? throop	[reply] [d/l]
Re^4: Hash initialization works, but why? by chromatic (Archbishop) on Jun 05, 2008 at 00:56 UTC
It gets more confusing (but then it gets more clear). @foo is an array Yes. while (1,2,3) is a list `1, 2, 3` is a list. The parentheses have nothing to do with its listishness. The parentheses are there only so that the parser doesn't parse the whole expression as: `(@foo = 1), 2, 3;` ... as it would if there were no grouping parentheses. They indicate which list you want to assign.	[reply] [d/l] [select]
Re^5: Hash initialization works, but why? by throop (Chaplain) on Jun 05, 2008 at 14:14 UTC
Re^4: Hash initialization works, but why? by ikegami (Patriarch) on Jun 04, 2008 at 23:31 UTC
List and Array have very clear and distinct definitions in Perl. List is a type of value. Array is a type of variable. You can't take the length of a list. There's no such thing as a reference to a list. There's no such thing as a list in scalar context. Come to think of it, that's in the FAQ.	[reply]
Re^4: Hash initialization works, but why? by ikegami (Patriarch) on Jun 05, 2008 at 02:38 UTC
One more tidbit. The variable `@foo` is an array, but the expression `@foo` evaluates to a scalar or a list, depending on context. `my $num_elemens = @foo; # @foo evaluates to a scalar. my ($first, $second) = @foo; # @foo evaluates to a list.` [download]	[reply] [d/l] [select]