Re: [RFC] Using gitrevisions :/search style with other operators

From: Jakub Narebski <jnareb@gmail.com>
Date: Nov 10, 2010, 15:26 UTC
Message-ID: <m3eiatfbg2.fsf@localhost.localdomain>
In-Reply-To: <20101110084609.78f532a5@chalon.bertin.fr>

Yann Dirson <dirson@bertin.fr> writes:

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> On Tue, 09 Nov 2010 23:46:59 -0800
> Kevin Ballard <kevin@sb.org> wrote:
>> On Nov 9, 2010, at 11:32 PM, Yann Dirson wrote:
>> 
>>>> Another thing to consider - the current :/foo syntax searches for
>>>> the newest commit reachable from any ref. Using the ^{} syntax will
>>>> require specifying a ref first. I'm not sure this is a problem
>>>> though, as I'm not really sure why :/foo searches from all refs to
>>>> begin with.
>>> 
>>> The syntax could be extended so that ^{whatever} starts looking at
>>> current commit (ie. HEAD), somewhat like @{whatever} looks at
>>> reflog for current branch.

The <ref>@<sth> is about reflogs: <ref>@{<n>}, <ref>@{<aproxidate>}, @{-<n>}, <ref>@{upstream} / <ref>@{u}. Because HEAD has separate reflog, then @{<sth>} is about current branch reflog (@{-<n>} uses HEAD reflog, though). <ref> must be something that has reflog.

The <obj>^<sth> is about dereferencing: <tag>^{<objtype>} and <tag>^{}, which includes following parents <commit>^ and <commit>^<n>. Then there are odd <commit-ish>^! (returning range) and <commit-ish>^@ (all parents) but which nevertheless follow this rule.

The <obj>:<sth> is (with single exception of ':/<regexp>') about selecting subitem (path): <tree-ish>:<path>, [:<stage>]:<path>

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>> 
>> :/foo doesn't start from the current commit - it searches all refs.
>> However, making ^{} search all refs if not given one doesn't make
>> sense for any operator except :/foo, so I don't think it's worth doing
> 
> Yes, that's why I suggested to make it search from HEAD, not from all
> refs.

Perhaps '--all^{/foo}'?  Just kidding... I think ;-)

About n-th match: I think that ^{<n>/foo} or ^{:<n>/foo}... or ^{:nth(<n>)/foo} as generic form of e.g. ^{3rd/foo} a la Perl 6 :-) Or perhaps even full form ^{m:nth(<n>)/foo}... well, perhaps not.

-- 
Jakub Narebski
Poland
ShadeHawk on #git