Dear monks,

I'm writing a method which removes the first argument of @_ if it is the blessed reference to the package or the class name / package, thus allowing several call syntaxes to be used for static methods (Not requiring any instance variables). I've had a quick super search but couldn't find anything.

Method example: ( Classic pow example ):

package Object;
{

    # pow does not require an object data
    sub pow 
    {
       my $val = shift;
       my $pow = shift;
   
       return $val ** $pow;
    }

}
1;
[download]

I can quite happily call this method like so:

use Object
Object::pow( 2, 2 );
[download]

or exported ( With some code added to the module):

use Object 'pow';
pow( 2,2 );
[download]

I thought it would be nice to call such method as any of the following:

Object->pow(2,2);

Object::pow(2,2);

my $obj = Object->new();
$obj->pow(2,2);

pow(2,2); # Exported or internal method call
[download]

To accomplish this I wrote a method which filters the arguments:

    sub rmvObjRef
    {
        my $package_name = shift;      
        my $param1       = ref( $_[ 0 ] ) || $_[ 0 ];

        # If param1 holds the name of the class / package
        # remove it from the arguments.
        if ( $param1 eq $package_name ) {
            shift @_;
        }

        return @_;

    }
[download]

or the one line equivalent to use inline:

shift if ref $_[0] eq __PACKAGE || $_[0] eq __PACKAGE__;
[download]

Which is called as follows.

@_ = rmvObjRef( __PACKAGE__, @_ );
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The only draw back to my approach I can see is if the arguments contain the object reference which the method has been declared in at $_[0]:

package NumObject;

sub addObjects
{
   @_ = rmvObjRef( __PACKAGE__, @_ );   

   # Just an example, Obviously an overloaded '+'
   # op could achieve this.
   return NumObject + NumObject;

}
[download]

Calling this as:

my $num1 = NumObject->new();
my $num2 = NumObject->new();
NumObject::addObjects( $num1, $num2  );
[download]

would remove $num1 from @_
Is there any way to just check just the left argument of the infix dereference operator "->"?

Many Thanks,
John,

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