Because a class can override isa to return what it wants, and you're bypassing that. 

package MyIO;

sub isa
{
  my $self = shift;
  my $type = shift;

  return 1 if $type eq 'IO::Handle';

  return $self->SUPER::isa($type);
}

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In this scenario, MyIO wants to claim to be an IO::Handle. That's perfectly allowed - presumably, then, it fulfills all the obligation of IO::Handle without actually deriving from it. Another example: 
package MyNotIO;

use base 'IO::Handle';

sub isa
{
  my $self = shift;
  my $type = shift;

  return 0 if $type eq 'IO::Handle';

  return $self->SUPER::isa($type);
}

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In this case, I don't want to look like an IO::Handle, even though I'm derived from it. Maybe I just want all the convenience of the IO::Handle type, while breaking its interface. Again, your UNIVERSAL::isa($obj, 'IO::Handle') would return incorrectly because it wouldn't be using this isa.

The problem with $fd->isa(...) is that it crashes if $fd is undef, whereas a simple "no" would be better. If they made that work at the same time as inheritance, that'd be awesome :-) Most likely, that would just be: 

package UNIVERSAL;

sub safe_isa
{
  my $obj = shift;
  return unless defined $obj and ref $obj;

  $obj->isa(@_);
}

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Now you could call UNIVERSAL::safe_isa($fd, 'IO::Handle'), and it'd work properly, and safely, even if $fd wasn't actually an object.

In reply to Re: Do not use UNIVERSAL::isa this way; but why? by Tanktalus
in thread Do not use UNIVERSAL::isa this way; but why? by klekker

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