Re: git fsck: unreachable vs. dangling

From: Junio C Hamano <gitster@pobox.com>
Date: Apr 14, 2015, 09:22 UTC
Message-ID: <xmqq1tjn3vzs.fsf@gitster.dls.corp.google.com>
In-Reply-To: <CAHGBnuPepR11r9qpxb2G7pGhq0iiaqCCLnbt5NB_xohHi=cVxA@mail.gmail.com>

Sebastian Schuberth <sschuberth@gmail.com> writes:

Show 19 quoted lines

> On Tue, Apr 14, 2015 at 10:50 AM, Michael J Gruber
> <git@drmicha.warpmail.net> wrote:
>
>> "to dangle" means "to hang loosely".
>>
>> So, in the description above, "A^ dangles from A loosely" because it
>> hangs from A (you can reach it from A) but loosely, because it would
>> "drop" if A gets dropped and A is "likely" to be dropped (because it is
>> unreachable by refs). But A^ is not dangling in our terminology.
>>
>> If you *reverse the arrows*, i.e. consider A^ pointing to A, it becomes
>> more apparent that A is dangling: it is an unreferenced leaf node.
>
> That's exactly what confused me. In the very literal sense, something
> can only "hang loosely", i.e. dangle, if it's only tied at *one* end,
> and that's the case for A (which is only connected to A^) but not for
> A^ (which is connected to its parent, and A). Especially when talking
> about A as a "leaf" node, like in the leaf of a natural tree, I would
> think that A is dangling.

I am not sure if I follow, but probably it is just me who is not strong at math, or whose eyesight is not keen enough to notice the arrow heads on links between the commits.

I just visualize commits to be ping-pong balls with strings between them, and then grab the root of the graph and lift the whole thing up, while tips of the branches and tags are anchored. Commit A will be dangling in the wind if you shake the whole thing.

But that visualization breaks down once you start thinking about what will happen to A^{tree} and its blobs; they are attached to A with thin strings and they will have to float above A (i.e. sit somewhere closer to the root of the tree) just like A^ will go closer to the root, to make A appear the "dangling" one, as the direction of the arrow is from A to A^{tree} just like we have an arrow from A to A^; just like A^ is not dangling because of A, A^{tree} is not dangling.