in reply to Overloading without infinite descent (fixed by time: see Perl 5.10.1)

package Before;
my %looper;
use overload
    '&{}' => sub {
        my ( $f ) = @_;
        my $fs = overload::StrVal($f);
        if( $looper{$fs} ){
            print "$fs   got looper already\n";
            return $looper{overload::StrVal($f)};
        } else {
            my $looper = 0;
            my $s = sub {
                $looper++;
                print "$fs  Stuff before (looper=$looper)\n";
                goto &$f unless $looper > 1;
            };
            $looper{$fs} = $s;
            return $s;
        }
    };

package main;
( bless sub {} => 'Before' )->()
__END__
Before=CODE(0x18305e4)  Stuff before (looper=1)
Before=CODE(0x18305e4)   got looper already
Before=CODE(0x18305e4)  Stuff before (looper=2)

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Re^2: Overloading without infinite descent
by JadeNB (Chaplain) on Aug 03, 2009 at 06:42 UTC
    This gives the advertised output, and certainly avoids the infinite descent, but it seems only to do what I want for the empty sub (or any other side-effect-free sub, I suppose)—if I execute ( bless sub { print "I am a side effect\n" } => 'Before' )->(), then I am a side effect is never printed.
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      Probably because that sub is never executed
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        Right, but my hope was that it would be—that is, that, upon de-referencing, the ‘pre-code’ would be executed first, then the code to which the reference actually points. Thus, I'd like to see: 
        ( bless sub { print "I am a side effect\n" } => 'Before' )->()

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        print 
        Stuff before
I am a side effect

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