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copied!<p>If you say <code>__PACKAGE__->run(@ARGV)</code> then <code>run</code> can be defined in a class you inherit from or allows a class to inherit from this one. If you just say <code>run(@ARGV)</code> you are missing the class information. This only matters if you are doing OO style programming.</p> <p>Put the following in a file named <code>Foo.pm</code>, and then say <code>perl Foo.pm 1 2 3</code></p> <pre><code>package Foo; use strict; use warnings; __PACKAGE__->main(@ARGV) unless caller; sub main { my $class = shift; my $obj = $class->new(@ARGV); print $obj->to_string, "\n"; } sub new { my $class = shift; return bless [@_], $class; } sub to_string { my $self = shift; return "args: " . join ", ", map { "[$_]" } @$self; } 1; </code></pre> <p>Now put the following in <code>Bar.pm</code>, and say <code>perl Bar.pm a b c</code>.</p> <pre><code>package Bar; use strict; use warnings; use base 'Foo'; __PACKAGE__->main(@ARGV) unless caller; sub to_string { my $self = shift; return "args: " . join ", ", map { "<$_>" } @$self; } 1; </code></pre> <p>Now lets see what happens if you don't use <code>__PACKAGE__</code> in this environment. Make the first section of code in <code>Foo.pm</code> look like this:</p> <pre><code>main(@ARGV) unless caller; sub main { my $obj = Foo->new(@ARGV); print $obj->to_string, "\n"; } </code></pre> <p>Now run <code>perl Foo.pm 1 2 3</code>. Everything should still look right. Now try running <code>perl Bar.pm a b c</code>. We still get output, but it is not the output we expect. What happens if we remove <code>__PACKAGE__</code> from <code>Bar.pm</code>? Well, we get and error: <code>Undefined subroutine &Bar::main called at Bar.pm line 8.</code> This is because there is no <code>main</code> function in the module and we are not calling it with the class, so it won't look for it in the <code>Foo</code> package anymore.</p>

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