FAQ 7.26 How can I find out my current or calling package?

Discussion in 'Perl Misc' started by PerlFAQ Server, Jan 22, 2011.

  1. This is an excerpt from the latest version perlfaq7.pod, which
    comes with the standard Perl distribution. These postings aim to
    reduce the number of repeated questions as well as allow the community
    to review and update the answers. The latest version of the complete
    perlfaq is at http://faq.perl.org .


    7.26: How can I find out my current or calling package?

    (contributed by brian d foy)

    To find the package you are currently in, use the special literal
    "__PACKAGE__", as documented in perldata. You can only use the special
    literals as separate tokens, so you can't interpolate them into strings
    like you can with variables:

    my $current_package = __PACKAGE__;
    print "I am in package $current_package\n";

    If you want to find the package calling your code, perhaps to give
    better diagnostics as "Carp" does, use the "caller" built-in:

    sub foo {
    my @args = ...;
    my( $package, $filename, $line ) = caller;

    print "I was called from package $package\n";

    By default, your program starts in package "main", so you should always
    be in some package unless someone uses the "package" built-in with no
    namespace. See the "package" entry in perlfunc for the details of empty

    This is different from finding out the package an object is blessed
    into, which might not be the current package. For that, use "blessed"
    from "Scalar::Util", part of the Standard Library since Perl 5.8:

    use Scalar::Util qw(blessed);
    my $object_package = blessed( $object );

    Most of the time, you shouldn't care what package an object is blessed
    into, however, as long as it claims to inherit from that class:

    my $is_right_class = eval { $object->isa( $package ) }; # true or false

    And, with Perl 5.10 and later, you don't have to check for an
    inheritance to see if the object can handle a role. For that, you can
    use "DOES", which comes from "UNIVERSAL":

    my $class_does_it = eval { $object->DOES( $role ) }; # true or false

    You can safely replace "isa" with "DOES" (although the converse is not


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    so please include as much information as possible and relevant in any
    corrections. The perlfaq-workers also don't have access to every
    operating system or platform, so please include relevant details for
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    PerlFAQ Server, Jan 22, 2011
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