use Very::Long::Module::Name as Foo;

[liz]

Why should Python have something that Perl doesn't have?

I wondered so myself after reading The Very::Long::Class::Name problem and all its responses.

broquaint's response (using Package::Alias' knowledge about aliasing stashes at compile time) and bart's response (which is the simple assignment of @ISA) made me come up with YAS (Yet Another Solution). Now I'm wondering whether I should upload this to CPAN or not.

SYNOPSIS

    use as;
    use Very::Long::Module::Name as Foo;
    my $foo = Foo->new;

package as;

# Make sure we have version info for this module
# Make sure we do everything by the book from now on

$VERSION = '0.01';
use strict;

# Make sure we can do a source filter

use Filter::Util::Call ();

# Satisfy -require-

1;

#---------------------------------------------------------------------
+------

# Methods needed by Perl

#---------------------------------------------------------------------
+------
#  IN: 1 class (ignored)

sub import {

# Initialize the fixer upper subroutine
#  Obtain parameters
#  Create the alias specification if appropriate
#  Return possibly adapted string

    my $fix = sub {
        my ($package,$rest) = @_;
        my $alias = $rest =~ s#\bas\s*((?:\w|\w::\w)+)## ?
         qq(BEGIN {*{"${1}::"} = \\*{"${package}::"}}) : ''; # Package
+::Alias approach
#         qq(BEGIN {\@${1}::ISA = '$package'}) : '';  # bart's version
        "${alias}use $package$rest";
    }; #$fix

# Install the filter as an anonymous sub
#  Initialize status

    Filter::Util::Call::filter_add( sub {
        my $status;

# If there are still lines to read
#  Convert the line if it appears to contain a use statement
# Return the status

        if (($status = Filter::Util::Call::filter_read()) > 0) {
            s#\buse\s+((?:\w|\w::\w)+)([^;]*)#$fix->( $1,$2 )#eg;
        }
        $status;
    } );
} #import
#---------------------------------------------------------------------
+------

__END__

=head1 NAME

as - load OO module under another name

=head1 SYNOPSIS

    use as;

    use Very::Long::Module::Name as Foo;

    my $foo = Foo->new;

=head1 DESCRIPTION

This module implements a source filter that allows using of modules wi
+th a
very long name to be used under another name.  This is basically achie
+ved
by creating a dummy module with the short name that inherits from the 
+module
with the large name and which therefore means that there is a run-time
penalty for this feature.

=head1 INSPIRATION

Inspired by bart's response (http://www.perlmonks.org/index.pl?node_id
+=299082)
to a thread about long module names on Perl Monks.

=head1 AUTHOR

Elizabeth Mattijsen, <[e-mail]>.

Please report bugs to <[e-mail]>.

=head1 COPYRIGHT

Copyright (c) 2003 Elizabeth Mattijsen <[e-mail]>. All rights
reserved.  This program is free software; you can redistribute it and/
+or
modify it under the same terms as Perl itself.

=cut

Comments welcome. Particularly about whether the stash alias approach, or the @ISA approach is best. And if you think this would be a waste of CPAN space ;-)

Liz

Update:
In this response I describe why this proposal, as well as any similarly functioning module such as Package::Alias, suffer from a serious flaw because they change the global namespace. Needless to say, I won;t be uploading this module to CPAN. Everybody can relax again now ;-)