From: Branden [mailto:[EMAIL PROTECTED]] > > try to define a method in package bar and try to call it > from $bar, like $bar->foo. Won't work either, you have > to ${$bar}->foo. Overloading should loose the magic > in the same sense that the method should not be called. No, $bar->asString and $baz->asString both work. They produce: Hello World > > package bar; > > @ISA = qw(foo); > > sub new { bless \my $key; \$key } > > You return a reference to the object... No, it returns an object reference... Try it. > Try printing $$bar, it will work... No, it doesn't. $$bar is an undefined scalar. sub new { my $ref = \ my $key; bless $ref; $ref } and sub new { bless \ my $key; \$key; } Both return an object reference. The former has magic, the latter does not. I'm sorry... I didn't mean to start an off-topic thread. Unless someone can find a way to shoe-horn in a discussion of language or internals magic handling for Perl6? Is there really no substantial documentation anywhere on magic? I suppose that is why it is called magic, eh? package foo; use overload '""' => 'asString'; sub asString { 'Hello World' } package bar; @ISA = qw(foo); sub new { bless \my $key; \$key } package baz; @ISA = qw(foo); sub new { my $ref = \ my $key; bless $ref; $ref } package main; my ($bar, $baz); $bar = bar::->new(); $baz = baz::->new(); print "\$bar $bar\n"; print "\$baz $baz\n"; print "\$bar->asString ", $bar->asString, "\n"; print "\$baz->asString ", $baz->asString, "\n"; print "\$\$bar ", $$bar, "\n"; Results In: $bar bar=SCALAR(0x1bbfc9c) $baz Hello World $bar->asString Hello World $baz->asString Hello World $$bar
