Rich Fromm <richard_fr...@yahoo.com> writes:
> Jeff King wrote
>> Fundamentally the problem is
>> that the --local transport is not safe from propagating corruption, and
>> should not be used if that's a requirement.
> I've read Jeff Mitchell's blog post, his update, relevant parts of the
> git-clone(1) man page, and a decent chunk of this thread, and I'm still not
> clear on one thing. Is the danger of `git clone --mirror` propagating
> corruption only true when using the --local option ?
If you use --local, that is equivalent to "cp -R". Your corruption
in the source will faithfully be byte-for-byte copied to the
destination. If you do not (and in your case you have two different
machines), unless you are using the long deprecated rsync transport
(which again is the same as "cp -R"), transport layer will notice
object corruption. See Jeff's analysis earlier in the thread.
If you are lucky (or unlucky, depending on how you look at it), the
corruption you have in your object store may affect objects that are
needed to check out the version at the tip of the history, and "git
checkout" that happens as the last step of cloning may loudly
complain, but that just means you can immediately notice the
breakage in that case. You may be unlucky and the corruption may
not affect objects that are needed to check out the tip. The initial
checkout will succeed as if nothing is wrong, but the corruption in
your object store is still there nevertheless. "git log -p --all"
or "git fsck" will certainly be unhappy.
The difference between --mirror and no --mirror is a red herring.
You may want to ask Jeff Mitchell to remove the mention of it; it
only adds to the confusion without helping users. If you made
byte-for-byte copy of corrupt repository, it wouldn't make any
difference if the first "checkout" notices it.
To be paranoid, you may want to set transfer.fsckObjects to true,
perhaps in your ~/.gitconfig.
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