Hello morgon,

I would have expected $y also to be 2, but instead the last element of the list is used instead...

But there is no list. I think what’s confusing here is that parentheses are used in Perl to make lists. But that only works in list context. In scalar context, as here, parentheses are only for grouping,¹ which in this case makes no difference:

13:20 >perl -wE "my $y = ('a', 'b'); say $y;"
Useless use of a constant ("a") in void context at -e line 1.
b

13:20 >perl -wE "my $y = 'a', 'b'; say $y;"
Useless use of a constant ("b") in void context at -e line 1.
a

13:20 >
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From perldata#List-value-constructors:

List values are denoted by separating individual values by commas (and enclosing the list in parentheses where precedence requires it):
        (LIST)
In a context not requiring a list value, the value of what appears to be a list literal is simply the value of the final element, as with the C comma operator.

Update: ¹Actually, parentheses are only used for grouping (precedence), regardless of context, as indicated in the perldata quote. My statement that parentheses are used to make lists is a little misleading: they are generally needed, but only because the assignment operator has higher precedence than the comma operator:

14:05 >perl -MData::Dump -wE "my @a = 't'; dd \@a;"
["t"]

14:13 >perl -MData::Dump -wE "my @a = 't', 'u'; dd \@a;"
Useless use of a constant ("u") in void context at -e line 1.
["t"]

14:13 >perl -MData::Dump -wE "my @a = ('t', 'u'); dd \@a;"
["t", "u"]

14:13 >
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Hope that helps,

Very cool.

I have always thought parentheses would create a list.

Oh well, a life time is probably not enough to master perl (a quite dubious compliment though...).

Hi

Is that intentional or documented somewhere (I am running 5.16.2)?

I think perlop, yup, see http://perldoc.perl.org/perlop.html#Comma-Operator

It certainly violates the principle of least surprise for me but maybe there is a good reason for this behaviour...

Not sure about the source of the surprise for you , previous experience with X language or something :) but it is a common one, a fun thread about it at If you believe in Lists in Scalar Context, Clap your Hands

use warnings;
use strict;

my $y = () = ("a", "b");
print "$y\n";

__END__ 
Prints:
2
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Hi morgon,

Personally I find the easiest way to think about this is by looking at the Comma Operator:

In scalar context it evaluates its left argument, throws that value away, then evaluates its right argument and returns that value. ... In list context, it's just the list argument separator, and inserts both its arguments into the list. These arguments are also evaluated from left to right.

And as Athanasius said, the parens are required because the assignment operator has higher precedence than the comma operator:

$ perl -MO=Deparse,-p -e ' $x= "a","b"  '
(($x = 'a'), '???');
$ perl -MO=Deparse,-p -e ' $x=("a","b") '
($x = ('???', 'b'));
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Hope this helps,
-- Hauke D