The only advantage of rule 2 is that there are no exceptions. When I have a difficult reference problem, I solve it with rule 2, then simplify it with the other rules, testing each change as I make it. You did motivate me to read the reference document again. I find that the text of the rule is very clear. They probably should say that it can be applied recursively. I agree that the explanation (especially the examples) is confusing. I have created a few examples which I hope are more helpful. Each example shows the same thing using the variable and then the equivalent reference. I have used the test module to prove that they are the same. 
use strict;
use warnings;
use Test::Simple tests => 6;

# Scalars
my $x = 3;

my $scalar_ref = \$x;
ok( $x eq ${$scalar_ref}, 'EXAMPLE I');

my $scalar_ref_ref = \$scalar_ref;
ok( $x eq ${${$scalar_ref_ref}}, 'EXAMPLE II' );    # Rule 2 recurrsiv
+e


# Array
my @array = (0, 1, 2, 3);

my $array_ref = \@array;
ok( join( @array) eq join( @{$array_ref} ), 'EXAMPLE III' );
ok( $array[3] eq ${$array_ref}[3], 'EXAMPLE IV' );  # value = 3

# Array-0f-Arrays
my @as = qw(a aa aaa aaaa);
my @bs = qw(b bb bbb bbbb);
my @cs = qw(c cc ccc cccc);
my @ds = qw(d dd ddd dddd);
my @AoA = (\@as, \@bs, \@cs, \@ds);

ok( join( @ds ) eq join( @{$AoA[3]} ), 'EXAMPLE V' );
ok( $ds[3] eq ${$AoA[3]}[3], 'EXAMPLE VI' );    # value = 'dddd'

[download]
Bill

In reply to Re^5: 5.26 sigil reference syntax in subfunction by BillKSmith
in thread 5.26 sigil reference syntax in subfunction by chenhonkhonk

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