My real-life code that inspired this question goes something like this:

my $count = my ($arg1, $arg2, $arg3) = split /\s+/, $argstring;
$count == 3 or die "3 arguments expected; you supplied $count\n";
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I can't say I care for this feature. Smells like premature optimization to me. And I notice it doesn't work across function calls. That could be a nasty surprise for someone someday.

my $count = my ($arg1, $arg2, $arg3) =
    split /\s+/, $argstring, -1;
#                          ^^^^
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- tye        

So if I want the real argument count, I have to split into a temporary array first, or else use a dummy trailing array in the assignment list. Gross.

Personally i think the issue here is that what you are trying to do is much less common than the opposite. Ie, more people would rather have split// be faster on spliting a few elements out scenarios than they would with have it be slower but more consistant with other list returning functions. After all, split may be quite the wrong way to do this. Arguably

my $count=()=$argstring=~/\s+/g;
my ($arg1, $arg2, $arg3) = split /\s+/, $argstring;
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is better anyway, as if the $argstring is long and you only are going to store three elements then its better not do the actual splitting (creating new svs, populating them etc) when you dont need to.

---
demerphq

my ($a,$b,$c);

my $n1 = ($a,$b,$c) = split /,/, "1,2,3,4,5";
my $n2 = my @x = ($a,$b,$c) = split /,/, "1,2,3,4,5";

print "$n1 ... $n2\n";
----
4 ... 3
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If it splits to one more, then why is that one going across one boundary, but not two?

Oh come off it.

sub list { return 1..5 }
my ($a,$b,$c);

my $n1 = ($a,$b,$c) = list();
my $n2 = my @x = ($a,$b,$c) = list();

print "$n1 ... $n2\n";
----
5 ... 3
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When you have scalar = list = list, you will always get the count of items in the rightmost list in scalar. In your $n2 example, your array is getting three items from the list. It isn't a scalar, so it isn't getting the count. What else would be expected? The behavior that merlyn is referring to is the reason why you don't see the complete count in the scalar.

So, based on what you're saying, $n2 should equal 5. Since @x isn't a scalar, it shouldn't affect the count of $n2 because I have scalar = list = list = list ... so, why doesn't that work?

Not to mention, where is it documented that When you have scalar = list = list, you will always get the count of items in the rightmost list in scalar. ?? I don't remember ever reading that ...

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You tricked yourself.

In the first case, the list assignment is evaluated in scalar context. In the second case, the list assignment is evaluated in list context and assigned to @x, yielding an array with three elements, which is then evaluated in scalar context. And an array with three elements evaluates to 3 in scalar context.

Makeshifts last the longest.