So, in answer to question 1: no, it contains a reference (which might be an address of something, but you don't {and shouldn't} need to know what).

Got it. Thanks for that clarification!

$ in Perl as a sigil means "give me a scalar value". In answer to 2: $$ means give me the scalar referred to by the scalar xxx. Because you can assign values to a scalar when you use $$ in an assignment you are assigning to the scalar referred to by the scalar. No pointers to be seen here.

Ok, I think I understand what you are saying. Taking your explanation one step further, I was wondering if I could use a "triple dollar sign" ($$$). Not that this would be idiomatic or anything but apparently you can:

my $variable = 22;
my $pointer = \$variable;

say "The address of \$variable, which contains the value $variable,";
say "is $pointer";

$$pointer = 25;

say "Look at that!  \$variable now equals $variable";

my $double_pointer = \$pointer;
$$$double_pointer = 50;

say "Look at that!  \$variable now equals $variable";
[download]

The address of $variable, which contains the value 22,
is SCALAR(0x801e64540)
Look at that!  $variable now equals 25
$double_pointer = REF(0x801e644b0)
Look at that!  $variable now equals 50
[download]

...Perl doesn't need "output parameters" because it can return multiple values from a sub.

I didn't know that! That is very cool!

References in Perl are most useful in building interesting structures which are a mixture of arrays, hashes and scalar values.

I guess I just need to keep re-reading the excellent documentation and trying this out. I implicitly understand the power of references but still struggling with applying them.