Re: Bug with rev-list --reverse?

From: Felipe Contreras <felipe.contreras@gmail.com>
Date: Apr 18, 2013, 11:15 UTC
Message-ID: <CAMP44s1np9vxZS6N2+w-K=zzRhq4x2+eFBzu-HJMeoNjk7ze-Q@mail.gmail.com>
In-Reply-To: <20130418102644.GY2278@serenity.lan>

On Thu, Apr 18, 2013 at 5:26 AM, John Keeping <john@keeping.me.uk> wrote:

Show 13 quoted lines

> On Thu, Apr 18, 2013 at 05:17:14AM -0500, Felipe Contreras wrote:
>> If I do these:
>>
>> % git log --oneline -1 v1.8.1.5^..v1.8.1.6
>> % git log --oneline --reverse -1 v1.8.1.5^..v1.8.1.6
>>
>> I expect to get a different output, and not both showing v1.8.1.6.
>> Wouldn't you agree?
>
> I expect to get the same output.  This is probably because I consider
> --reverse to be an output filter.  So I expect to show the commits
> "v1.8.1.5^..v1.8.1.6 -1" which selects a single commit and then show
> that in reverse order.

How about this:

% git log --oneline --reverse --max-count=1 v1.8.1.5^..v1.8.1.6

In this case --max-count is acting as "start from the first commit before the tip", not as "output a maximum of one commit". Given that the name is max-count, I expect it to be the later.

And if max-count doesn't select a maximum of n commits, then what does?

-- 
Felipe Contreras