Are there cases where something that's unambiguously a list (and not an array) returns anything else than the last item?

A regular expression match with the /g flag, in scalar context.

A map or grep expression in scalar context.

The empty list my $x = () = some_expression; construct, in scalar context.

split, chomp, ....

A bare hash. (Want to argue over what a list is? Is it multiple, comma-separated expressions or something that pushes one or more items onto the internal stack? I'm not sure lists even exist as a language-level construct in Perl 5.)

One could argue that built-ins like map and grep don't return a list in scalar context, but are context sensitives themselves, so they are not fit for explaining "a list in scalar context". Likewise a regex match with /g.

Sure, but you can also argue that 1, 2, 3, isn't a list of values but a list of expressions. Is this a list:

my $x = ($y = 1), ($z = 2);

How about this?

s{\\}{\\\\}g, print

Or this?

foo => scalar param( 'foo' )

It's difficult to tell without seeing the context... which makes me wonder if lists do exist, or if we should talk about the value of expressions instead.

chomp???   How did chomp get in there?   It never returns a list.

my @x = chomp <DATA>;

Bingo, @x is gets a one-element list -- or at least, it's a list in list context and a scalar in scalar context. Tricky! Here's a mind-bender. What's this?

1, 2, 3

It could be a three-element list, in list context, or it could be a scalar and two void expressions (which get optimized away), in scalar context. In void context, it could be nothing.

Update: Reworded one sentence.

when used with a list, it returns the total number of characters chomped... and not a list (so, you are right that there is no "list context" to it.
[]s, HTH, Massa (κς,πμ,πλ)

On $r = () = 4..5; vs @r = () = 4..5;

It's useful to look at this snippet to help understand what's happening:

$r = ($x, $y) = (1..10);
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• Assignment is right associative.
• ($x,$y) = 1..10 sets $x, and $y and returns in scalar context returns the number of elements on the right hand side (in this case, 10). In a list context it will return a list of lvalues assigned to.
• So now we have reduced the above to $r = 10 which needs no explanation.
• Note that it doesn't matter what is in the middle term, as long as it puts the right hand assignment in list context.

If we look at @r = () = 1..10 we will see that the right hand assignment assigns to no lvalues and so returns an empty list. And thus @r is empty.

Perhaps I don't fully understand right associativity.

Consider:

$r = () = 0..11 ;
$s = ($x, $y) = 0..11 ;
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which I thought meant that the 0..11 would be evaluated in List Context, and then assigned to the lvalues in lists () and ($x, $y) -- the rightmost assignment first.  Then the result of that assignment would be assigned to <c>$r and $s this being a scalar assignment. The actual result is that both $r and $s are set to 12. This seems to imply that the first, list, assignment assigns the entirity of (0, 1, ...11) to the intermediate list, irrespective of the number of lvalues it contains. Then the length of that list is assigned to $r. This last step is consistent with the well known Array in Scalar Context... though () doesn't look like an array to me... but I'm willing to believe.

Of course:

$s = ($x, $y) ;
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assigns $y to $s. But I'm willing to accept that this is quite different from the above. I just don't yet see how.

As you say, if we then look at:

@r = () = 0..11 ;
@s = ($x, $y) = 0..11 ;
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we find that @r is set to the empty list and @s to the two element list (0, 1). I confess this is more what I expected. The result of the second assignment depends on the result of the first (rightmost).

Is it just me ? Is this in fact consistent at some deeper level I don't yet understand ?

I think the inner consistency you are missing is that operands aren't only values, they can be operations as well. The left = (the scalar assignment) has as its right operand not the "result of [the list] assignment", but rather the list assigment operation. And that operation, by virtue of being the right side of a scalar assignment, gets scalar context, and the "result" of a list assignment in scalar context is the number of elements on the right of the assignment. Does that help at all?
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The difference between

$s1 = ($x, $y) = 0..11;
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and

$s2 = ($x, $y)
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is that $s1 gets assigned the result of the list assignment operator, and $s2 gets assigned the result of the scalar comma operator. In neither case there's a list assigned to any of the $s? variables; in both cases the result of operators are assigned.

But as long as you keep believing in the existence of lists in scalar context, you'll keep being confused.

The only exception I'm aware of is $r = () = 4..5, where the number of items is returned rather than the the last item. But by my definition above that would actually be an array, not a list, because you can assign to it.