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

Hi monks,

I have one question about passing subroutines as arguments in packages.

Here is my simple code for question:

* file: FooConfig.pm 

package FooConfig;

use strict;

[ constructor and some functions ]

sub important_func
{
 my $self = shift;
 my $arg = shift;

 my %hash =
 (
  '1' => 'red',
  '2' => 'blue',
  '3' => 'green',
 )

 return $hash{ $arg } if $hash{ $arg };
}

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* file: Foo.pm

package Foo;

use strict;

[ constructor and some functions ]

sub somesub_foo
{
 my $self = shift;

 [...]
 use FooConfig;
 my $FooConfig = FooConfig->new;

 my $ref = \&FooConfig->important_func( '1' );

 [ do something with this ]
}

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Result is: Undefined subroutine &Foo::FooConfig->important_func called at /path_somewhere/Foo.pm line 123

Im trying &$FooConfig, \$FooConfig and many other this without success. :(

Valid usage this function: $FooConfig->important_func, but reference return &PackageName:: before and i got this error. What's wrong with this?

Best regards,

Replies are listed 'Best First'.
Re: Passing subroutine as argument between packages
by Eliya (Vicar) on Nov 30, 2011 at 14:26 UTC
    Valid usage this function: $FooConfig->important_func, but reference return &PackageName:: before

    I'm not sure I understand what your problem is.  As it seems, you appear to have figured out that the correct way to call the method would be $FooConfig->important_func( ... ). So where/when are you getting the &PackageName:: you don't want?

    Upon rethinking, maybe you want something like this 

        my $ref = sub { $FooConfig->important_func( '1' ) };

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    which would 'bind' the object instance and the method parameter to the coderef, so that you could call it later on as $ref->().

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      Thank you for you answer,

      I want ask you about one more thing: in this method, how pass eventually arguments for subroutine $FooConfig->important_func?

      I checked $ref()->('a' ), $ref()('a'), $ref->( 'a' ), $ref->()('a'), $ref->()->('a') or like this, and didn't work.

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[d/l]

        Given:

        my $ref = sub { $FooConfig->important_func( '1', @_ ) };

        ... then you can write:

        $ref->( 'more', $args, %here );

        ... though be cautious and understand how the function reference passes its arguments to the inner function.

        Improve your skills with Modern Perl: the free book.

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