I've seen this said before, but

P:\test>perl -MO=Deparse -e"@a = 1,2,3,4,5; print @a"
@a = 1, '???', '???', '???', '???';
print @a;
-e syntax OK

P:\test>perl -e"@a = 1,2,3,4,5; print @a"
1


P:\test>perl -MO=Deparse -e"@a = (1,2,3,4,5); print @a"
@a = (1, 2, 3, 4, 5);
print @a;
-e syntax OK

P:\test>perl -e"@a = (1,2,3,4,5); print @a"
12345
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Empirically, that suggests that the parens are, at least in some cases, required to build a list.

But they don't create lists. They are 'required' just as they are in some cases required to call a function correctly. That doesn't mean parens call functions. In the case you describe, parentheses are used for precedence:

perl -MO=Deparse,-p -e"@a = 1,2,3,4,5; print @a"
((@a = 1), '???', '???', '???', '???');
print(@a);
-e syntax OK
[download]

A list is still created - it's just not the list you want.

Which goes to demonstrate that saying that , and => builds lists is not enough.

Just as parens are often required to clarify the programmers intent, so mentioning parens, and precedence is required to clarify the language's definition of a list.

---

Examine what is said, not who speaks.

"Efficiency is intelligent laziness." -David Dunham
"Think for yourself!" - Abigail
"Memory, processor, disk in that order on the hardware side. Algorithm, algorithm, algorithm on the code side." - tachyon

sub foo {
    $c = wantarray;   
    print $c ? "A" : defined $c ? "S" : "V"
}

     (foo(), foo(), foo()); print "\n";   
$v = (foo(), foo(), foo()); print "\n";   
@v = (foo(), foo(), foo()); print "\n";   
__END__
VVV
VVS
AAA
[download]

Well, no. That's a case of a list in scalar or list context. I quote from perlop about the assignment operator.

a list assignment in list context produces the list of lvalues assigned to, and a list assignment in scalar context returns the number of elements produced by the expression on the right hand side of the assignment.