Ahh, I see why you're confused. It's context again.

Arguments passed to a subroutine are passed as a list. If you put in a hash, it'll be flattened into a list:

my %hash = (
    one => 1,
    two => 2,
);

foo(%hash);

sub foo {
   print join(' | ', @_), "\n";
}
[download]

You can coerce these arguments into a hash inside your subroutine. It's not often done, though, because it's inefficient and prone to weird warnings if you don't have an even-sized list. Worse yet, if you want to pass a list and a hash, or a scalar and a hash, or anything else, you can't disambiguate between them:

sub foo {
    my %hash = @_;
}

# works fine
foo(%hash);

# uh oh
foo('nameofhash', %hash);

# not good at all
foo(@names, %hash);
[download]

If you can coerce all of your arguments into scalars, it's a lot easier to deal with them. (It's also more efficient, if you have a large list.) That's one reason people pass things by reference:

# pass a reference to %hash, which is a scalar
foo(\%hash);

sub foo {
    my $hashref = shift;
}
[download]

You'll have to dereference it if you want to treat it like a hash, but that's usually a small price to pay. Since the hash reference is already a scalar, you can pass it to other subs as is:

sub foo {
    my $hashref = shift;
    # add data
    bar($hashref);
}

sub bar {
    my $hashref = shift;
    # manipulate data as above
}
[download]

A reference to a hash is already a scalar (that's the point), so it's not necessary to take another reference to it. You'll just end up balding, with lots of curly braces. :)

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