> 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.lesp...@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: Acked-by: Stephen Finucane <stephen.finuc...@intel.com> _______________________________________________ Patchwork mailing list Patchwork@lists.ozlabs.org https://lists.ozlabs.org/listinfo/patchwork