It has nothing to do with functions...

\() returns a list of references to the variables in the list, so in this case a list of three references to the scalars. (Think (\"one",\"two",\"three")). But you assign it to a scalar... (So it becomes the last entry the list, which is 'three')

Example 2's $ref2 is just a reference to @list, so it indeed will return ARRAY when you ask what is refers to...

And Example 3's \() will give you a list of references to what's in the parentheses. But @list in list context, gets it's list expanded, so it's a list of three scalars again and assigning it to a scalar will give you the last entry 'three' again, thus ref $ref3 still is 'SCALAR'.

use Data::Dumper;
print Dumper($ref1), "\n";
print Dumper($ref2), "\n";
print Dumper($ref3), "\n";

[download]
Should clear things up...

Update
In case of the $n = qw(one two three) makes $n eq 'three' not being clear, refer to chapter 2.3.4 of Programming Perl... It boils down to the list 'knowing' that it's being used in scalar context, and thus evaluating to the value of the last element, just like in C.

An array is slightly different from a list; an array returns it's size when evaluated in scalar context. Only in example 2 do you make the list an array by assigning it to @list.

In reply to Re: \ Operator in referencing by Gilimanjaro
in thread \ Operator in referencing by dsb

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