$bin = sprintf "%b", oct "0x$hex";
$hex = sprintf "%x", oct "0b$bin";
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my $bin = sprintf "%b", hex $hex;
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sub bin2dec 
{ 
   my @bits = reverse split '', shift; 
   my $num = 0; 
   for( my $i=0; $#bits >= $i; ++$i )
   { 
      $num += 2 ** $i if $bits[$i] 
   } 
   return $num 
}

my $hex = sprintf "%x", bin2dec( $bin )
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That's not bin2dec, that's bin2num, unless you're running Perl on a machine that is using BCD internally.

Perl handles "num2dec" and "dec2num" so trivially that people sometimes get confused into believing that the numbers are stored as decimal internally. They're not.

-- Randal L. Schwartz, Perl hacker

Yes. Of course, Perl programmers enjoy that level of abstraction. Not worrying (too much) about how things are really stored makes it easier to focus on the important stuff... like algorithms. Thanks for your comment though.

Since you seem to have bin2num, here's bin2dec, taken straight from the pack() docs:

sub bin2dec {
  unpack("N", pack("B32", substr("0" x 32 . shift, -32)));
}
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-OzzyOsbourne