Re: [PATCH v3 1/2] revision: differentiate if --no-abbrev asked explicitly

From: Đoàn Trần Công Danh <congdanhqx@gmail.com>
Date: Aug 14, 2020, 14:50 UTC
Message-ID: <20200814145043.GB9518@danh.dev>
In-Reply-To: <xmqqzh6yyr1t.fsf@gitster.c.googlers.com>

On 2020-08-13 18:06:22-0700, Junio C Hamano <gitster@pobox.com> wrote:

Show 26 quoted lines

> Đoàn Trần Công Danh  <congdanhqx@gmail.com> writes:
> 
> > Let me list some combination here:
> >
> > * none of --abbrev --no-abbrev --full-index -> default short index
> > * --abbrev --full-index                     -> full-index
> > * --full-index --abbrev                     -> full-index
> > * --abbrev --no-abbrev                      -> full-index
> > * --no-abbrev --abbrev=[n]                  -> shortened index to n char
> >
> > So, we can't use full_index bit, because --no-abbrev can be defeated
> > by --abbrev, but --full-index will always win --abbrev.
> 
> Sure, I wasn't suggesting to flip the flags.full_index bit upon
> seeing "--no-abbrev".  When --no-abbrev is in effect (i.e. the last
> one among --no-abbrev, --abbrev, or --abbrev=n), .abbrev field is
> set to 0.  So wouldn't it be sufficient to say
> 
>  - If flags.full_index bit is set, show the full object name
> 
>  - If abbrev is 0, show the full object name
> 
>  - All other cases, after clamping the value of abbrev to reasonable
>    value, truncat the object name to that length
> 
> What am I missing?

No, you didn't miss anything.

It's obviously me, who screwed up the logical thinking.

Originally, I come up with something along the line in 2/2:

	int abbrev = o->flags.full_index ? hexsz : DEFAULT_ABBREV;
	if (!o->abbrev)
		abbrev = o->abbrev;

I couldn't recalled what I wrote, but that logic requires 1/2, after I come up with 1/2, I re-analysed 2/2 and come up with current logic.

I failed to re-visit 1/2 to check if it's necessary. It's all MY fault.

Sorry for wasting everyone's time in 1/2.

Please eject 1/2 from this series.

-- 
Thanks

Danh