Re^3: Hash initialization works, but why?

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)
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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

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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
"Parentheses...have nothing...to do with...its listishness." Ah. Understanding is slowly dawning to this refugee from 10 years of LISP programming. throop	[reply]
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]