jaa has asked for the wisdom of the Perl Monks concerning the following question:

Dear Monks

Is there a special var or some other way that tells me the filename of the current module?

I want to do something like: 

  my $package = __PACKAGE_FILENAME__;
  my $related = $package;
  $related =~ s/this.pm/other.pm/;
  if ( -e $related ) {
    require $related;
  else {
    die "related package missing: $related";
  }

[download]

Thanks!

Jeff

Replies are listed 'Best First'.
Re: Filename of current package
by amw1 (Friar) on Mar 22, 2004 at 18:39 UTC
    caller may give you what you want.
    foo.pl 
    use Data::Dumper;

sub callstack
{
 return caller();
}

print Dumper(callstack());

    [download]
    This prints 
    $VAR1 = 'main';
$VAR2 = 'foo.pl';
$VAR3 = 9;

    [download]
    Var1 is the calling package (in this case main) Var2 is the calling file Var3 is the line where it was called
[reply]
[d/l]
[select]

      Thanks for the pointer, but I'm still not able to find the filename of the package. I have played a little with caller(), but it seems to provide details of the calling program, rather than the included module?

      Here is more what I am looking to do: 

      # my.pl
use My::Factory;

my $obj = My::Factory->new('Apple');
...
# My/Factory.pm
package My::Factory;
1;
sub new{
  my $factoryclass = shift;
  my $class = "My::" . shift; # My::Apple
  my $filename = $class;
  $filename =~ s/::/\//g;  
  $filename .= '.pm'; # My/Apple.pm
  if ( -e $filename ) {
    require $filename;
  } else {
    die "not found: $filename";
  }     
}

      [download]

      Which all looks good and dandy, except that My::Factory is picked up from @INC at runtime. I thought that a package might know its own filename?

      Note that I don't want to pick up My::Apple from somewhere else in @INC - I want it to be in the same place as My/Factory.pm

      Apologies if that is not clear...

      Off to see if I can figure out how Carp does it...

      Regards

      Jeff

[reply]
[d/l]
Re: Filename of current package
by ambrus (Abbot) on Mar 22, 2004 at 20:08 UTC

    See perldoc perldata:

    The special literals __FILE__, __LINE__, and __PACKAGE__ represent the current filename, line number, and package name at that point in your program. They may be used only as separate tokens; they will not be interpolated into strings. If there is no current package (due to an empty "package;" directive), __PACKAGE__ is the undefined value.

[reply]
      Even better than caller() - thank you very much! +++
[reply]
Re: Filename of current package
by jaa (Friar) on Mar 22, 2004 at 19:38 UTC

    amwl - thank you, caller() is the key, and Carp::Heavy has the detail - my problem was that I was doing it in the new() - checking caller in a sub-routine of the same package provides the answer: 

    
package My::Factory;
1;
sub new {
  my $factory = shift;
  my $name    = shift;
  my $classfile = package_name();
  $classfile =~ s/Factory.pm$/$name.pm/;
  if ( -e $classfile ) {
    require $classfile;
    my $classname = "My::$name";
    return $classname->new(@_);
  } else {
    die "class not found: $classfile";
  }
}

sub package_name {
  my (@called) = caller(0);
  return $called[1];
}

    [download]
[reply]
[d/l]

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