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Foo is not a Bar, why?

by liz (Monsignor)
on Dec 18, 2003 at 15:23 UTC ( #315535=perlmeditation: print w/ replies, xml ) Need Help??

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  1. or download this
    my $foo = Foo->new;
    print "foo is".($foo->isa( 'Bar' ) ? '' : ' _not_')." a Bar\n";
    ...
    sub new { bless {},shift }
    __END__
    $foo is _not_ a Bar
    

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