That's a scalar that contains a reference to a hash.

Observe (untested):

my %hash = (
    a => 1,
    b => 2,
);

my $href = \%hash;

[download]

The coder just skipped the intermediary step of taking a reference to the hash, and instead constructed it as a hash reference directly. To get information from the hash as I depicted above, you access it normally:

my $thing = $hash{a};

[download]

In the case of the hash reference, you have to dereference first:

my $thing = $href->{a};

[download]

You can dereference the entire hash reference and put it into a new hash if you please, but there's not often reason to do so unless you need a copy of the entire structure:

my %hash = %{ $href };

[download]

See perlreftut.

In reply to Re: Confused on Scalar/Hash by stevieb
in thread Confused on Scalar/Hash by csharp2a

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