Re: A better mod (%) operator?

[fglock]

This is something I used in the old times when multiplication was much faster than division:

sub divmod {
  use integer;
  my ($a, $b) = @_;
  my $tmp = $a / $b;
  ($tmp, $a - ($b * $tmp) );
}

$a = 13;
$b = 2;

print divmod($a, $b);

This is much more interesting to do in assembler then in Perl (maybe Parrot?)