On Thursday 17 July 2008 19:15:52 James Keenan via RT wrote:
> + if (! defined $args->{revision}) {
> + $args->{revision} = 'unknown';
> + _print_to_cache($args->{cache}, $args->{revision});
> + return $args->{revision};
> + } else {
> + if (defined ($args->{prev}) && ($args->{revision} ne
> $args->{prev})) { + _print_to_cache($args->{cache},
> $args->{revision});
> + return $args->{revision};
> + }
> + else {
> + return $args->{current};
> + }
> }
> }
I would find this clearer if you reversed the top level conditions. In
English, this reads:
if the revision isn't defined, then do this
otherwise do that
Flipping the order changes it to:
if the revision is defined, do this
otherwise, do that
There's no double-negative.
-- c
