perlref states that there is a difference between taking a reference to a list, and a reference to an array:
Taking a reference to an enumerated list is not the same
as using square brackets--instead it's the same as 
creating a list of references! 

    @list = (\$a, \@b, \%c);
    @list = \($a, @b, %c);	# same thing!  
 
As a special case, \(@foo) returns a list of references to
the contents of @foo, not a reference to @foo itself. 
Likewise for %foo, except that the key references are to 
copies (since the keys are just strings rather than full-
fledged scalars). 
So I have two questions:
  1. Why the distinction? Why are there two seperate behaviors?
  2. What is happening that creates the difference in behavior?
#!/usr/bin/perl use strict; # example 1 my $ref1 = \(qw(one two three)); # should be called in list context # example 2 my @list = qw(one two three); my $ref2 = \@list; # example 3 # returns same as example 1 my $ref3 = \(@list); # should be called in list context print ref $ref1, "\n"; print ref $ref2, "\n"; print ref $ref3, "\n";
Based on these examples, my best guess is that the parentheses are the key. Going back to the "if it looks like a function, then it is a function" philosophy behind the interpreter, I guess \() does look like a function. As such, \("one","two","three") and \(@list) gets executed like a function call with n arguments (3 in the case of the list). @list gets flattened like any other list passed as an argument and so the behavior is basically the same for both cases.

That's my best guess anyway...


dsb
This @ISA my cool %SIG

In reply to \ Operator in referencing by dsb

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