What does __PACKAGE__->{foo} mean ?

Question

I'm refactoring a perl module in legacy code, and this is a function from the module:

sub get_user {
    my $user = __PACKAGE__->{user};
    if (!defined $user) {
       # more code
       __PACKAGE__->{user} = $user;
    }
    return $user;
}

This module compiles under use strict. And there's no package variables defined. What does __PACKAGE__->{user} mean?

Answer 1

__PACKAGE__->{user} = $user sets the value of user in whatever the current package is.

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From perldoc perlmod:

The special symbol __PACKAGE__ contains the current package, but cannot (easily) be used to construct variable names.

Answer 2

__PACKAGE__ is a hash. This syntax accesses a keyed value.

Answer 3

__PACKAGE__ is the name of the current package; your code is using it as a symbolic hash reference. So if your package is foo, it is setting $foo::foo{'user'}. This is kind of a strange thing to do; I suspect it may be an error.

Because it's a symbolic reference, it shouldn't be allowed under strict. It seems to be, however, at least when the current package has multiple parts (e.g. Foo::Bar, not just Foo). I wouldn't depend on this bug staying in effect, though.