On Wed, Oct 3, 2012 at 7:25 AM, Angelo Borsotti
<angelo.borso...@gmail.com> wrote:
> Hi Phil,
>
>> Perhaps the confusion arises from the the meaning of "the safety".  In
>> this case, the safety mechanism in place is to prevent you from
>> creating a child commit which has the same "tree" contents (working
>> directory) as the parent commit.  It will not be the same commit
>> because it has different parent(s) than its parent commit; but the
>> tree (working directory) is the same and git normally prevents you
>> from doing this because normally this is an accident, a mistake.
>>
>> --allow-empty tells git you intend to do this and so it should bypass
>> this "no changed files" safety mechanism.  It is not a safety to
>> prevent you creating a new commit with the exact same sha1; the safety
>> is concerned only with the exact same "working directory" file
>> contents.
>>
>> Can you suggest a rewrite of this description which would make it more clear?
>
> Instead of:
>
> "Usually recording a commit that has the exact same tree as its sole
> parent commit is a mistake, and the command prevents you from making
> such a commit. This option bypasses the safety, and is primarily for
> use by foreign SCM interface scripts."
>
> I would suggest:
>
> "Usually recording a commit that has the exact same tree as its sole
> parent commit is not allowed, and the command prevents you from making
> such a commit. This option allows to disregard this condition, thereby
> making a commit even when the trees are the same. Note that when the
> tree, author, parents, message and date (with the precision of one
> second) are the same as those of an existing commit object, no new
> commit object is created, and the identity of the existing one is
> returned."

But that's true of 'git commit' generally; it has nothing to do with
--allow-empty.

-PJ

Gehm's Corollary to Clark's Law: Any technology distinguishable from
magic is insufficiently advanced.
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