I suppose they could have done that, but I can't comment on why they didn't. It would have to be able to detect whether it was working on a filehandle created by open or a dir handle created by opendir, but I guess that's not unreasonable.

Excuse me for keeping asking questions about this :) but it's for my own curiosity :

What could be the different ways to detect the kind of filehandle, to know if it was opened by open or opendir ?
Apparently, the following quote :

"DIRHANDLEs have their own namespace separate from FILEHANDLEs." (source)

could help detecting whether it was opened by open or opendir.
But when I do something like the following code to see the difference between the two namespaces, I get the same result for both of them.

open(DIR, './random_file');
print 'open : ', *main::DIR{NAME}, ' - ', *main::DIR{PACKAGE}, "\n";

opendir (DIR, '.');
print 'opendir : ', *main::DIR{NAME}, ' - ', *main::DIR{PACKAGE}, "\n"
+;
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Am I doing something wrong, or is there another way to differentiate the namespaces of the handles ?
Or could it be the version of Perl I'm using ( 5.005 ) ?

Thanks

Orthagonality with the C opendir(2) and friends would be my guess. If it really mattered to you you probably could implement a tied filehandle that provided its own map from readline to readdir without too much trouble.

Update: Yeah, not too much trouble at all.

#!/usr/bin/perl
use strict;

package Tie::Dirfile;
use IO::Dir ();

sub TIEHANDLE {
  my $class = shift;
  my $self = {};

  $self->{_dir} = shift;
  $self->{_handle} = IO::Dir->new( $self->{_dir} );
  
  return bless $self, $class;
}

sub READLINE { return shift()->{_handle}->read; }
sub CLOSE { return shift()->{_handle}->close( ); }

package main;

local( *DH );
tie *DH, 'Tie::Dirfile' => "/bin"
  or die "Can't create Tie::Dirfile for /bin: $!\n";

my $i = 0;
print $i++, ": ", $_, "\n" while( <DH> );

close( DH );

exit 0;

__END__
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