Re: Concurrent pushes updating the same ref

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On 11-01-06 02:37 PM, Junio C Hamano wrote:
> Jeff King <peff@xxxxxxxx> writes:
> 
>> On Thu, Jan 06, 2011 at 10:46:38AM -0500, Marc Branchaud wrote:
>>
>>> fatal: Unable to create
>>> '/usr/xiplink/git/public/Main.git/refs/builds/3.3.0-3.lock': File exists.
>>> If no other git process is currently running, this probably means a
>>> git process crashed in this repository earlier. Make sure no other git
>>> process is running and remove the file manually to continue.
>>> fatal: The remote end hung up unexpectedly
>>>
>>> I think the cause is pretty obvious, and in a normal interactive situation
>>> the solution would be to simply try again.  But in a script trying again
>>> isn't so straightforward.
>>>
>>> So I'm wondering if there's any sense or desire to make git a little more
>>> flexible here.  Maybe teach it to wait and try again once or twice when it
>>> sees a lock file.  I presume that normally a ref lock file should disappear
>>> pretty quickly, so there shouldn't be a need to wait very long.
>>
>> Yeah, we probably should try again. The simplest possible (and untested)
>> patch is below. However, a few caveats:
>>
>>   1. This patch unconditionally retries for all lock files. Do all
>>      callers want that?
> 
> I actually have to say that _no_ caller should want this.  If somebody
> earlier crashed, we would want to know about it (and how).  If somebody
> else alive is actively holding a lock, why not make it the responsibility
> of a calling script to decide if it wants to retry itself or perhaps
> decide to do something else?

I'm not sure I follow this.

How would retrying a few times prevent us from finding out about an earlier
crash?  It's not like we're overriding the lock by retrying.  Nobody's going
to be able to remove a lock created by a crashed process, right?

And if someone active doesn't release the lock and the low-level code retried
a few times, the caller can still decide what to do.  I don't see how it
would even impact that decision -- if the caller wants to try again, the
system can still retry a few times underneath the caller's one retry.  It
seems fine to me.

		M.
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