[PATCH 2/2] retag: Properly display the final count

[Damien Lespiau]

On Mon, Nov 09, 2015 at 02:59:49PM +0000, Finucane, Stephen wrote:
> > On Mon, Nov 09, 2015 at 02:37:54AM +0000, Finucane, Stephen wrote:
> > > > i == count cannot happen in the loop as i will vary from 0 to count -
> > 1.
> > > >
> > > > Signed-off-by: Damien Lespiau <damien.lespiau at intel.com>
> > > > ---
> > > >  patchwork/management/commands/retag.py | 3 ++-
> > > >  1 file changed, 2 insertions(+), 1 deletion(-)
> > > >
> > > > diff --git a/patchwork/management/commands/retag.py
> > > > b/patchwork/management/commands/retag.py
> > > > index e67d099..cb95398 100644
> > > > --- a/patchwork/management/commands/retag.py
> > > > +++ b/patchwork/management/commands/retag.py
> > > > @@ -38,7 +38,8 @@ class Command(BaseCommand):
> > > >
> > > >          for i, patch in enumerate(query.iterator()):
> > > >              patch.refresh_tag_counts()
> > > > -            if (i % 10) == 0 or i == count:
> > > > +            if (i % 10) == 0:
> > >
> > > What happens if count == 11?
> > 
> > I don't where you are going with this question. In the loop we'll print
> > something out when i is 0, then 10. Out of the loop we'll print out
> > 11/11. But that cannot be what you were asking, am I missing a question
> > behind the question?
> 
> That's what I get for trying to be less prescriptive heh. It doesn't help
> when I give the wrong value also. I meant what happens if count == 12? From
> what I can see, it should have looked like this:
> 
>     if (i % 10) == 0 or i == count - 1:
> 
> i.e. every 10th iteration, or on the last iteration. However, the presence of
> the print on the line after this loop handles that last iteration case making
> this statement, as you say, unnecessary. Therefore:

That wouldn't work either. On the last iteration (count - 1)/count would
be printed and count/count would never be printed.

-- 
Damien