On 4/6/2017 6:48 PM, Stefan Beller wrote:
On Thu, Apr 6, 2017 at 1:37 PM, <git@xxxxxxxxxxxxxxxxx> wrote:
From: Jeff Hostetler <jeffhost@xxxxxxxxxxxxx>
Teach traverse_trees_recursive() to not do redundant ODB
lookups when both directories refer to the same OID.
And the reason for this is that omitting the second lookup
saves time, i.e. a lookup in the ODB of a sufficiently large
repo is slow.
My kneejerk line of thinking:
* yes, it sounds good to reduce the number of ODB accesses.
* But if we consider ODB lookups to be slow and we perform
a structured access, how about a cache in front of the ODB?
* We already have that! (sort of..) 9a414486d9 (lookup_object:
prioritize recently found objects, 2013-05-01)
* Instead of improving the caching, maybe change the
size of the problem: We could keep the objects of different types
in different hash-tables.
object.c has its own hash table, I presume for historical and
performance reasons, this would be split up to multiple hash
tables.
Additionally to "object *lookup_object(*sha1)", we'd have
a function
"object *lookup_object(*sha1, enum object_type hint)"
which looks into the correct the hash table.
If you were to call just lookup_object with no hint, then you'd
look into all the different tables (I guess there is a preferrable
order in which to look, haven't thought about that).
In operations such as read-tree, checkout, and merge when
the differences between the commits are relatively small,
there will likely be many directories that have the same
SHA-1. In these cases we can avoid hitting the ODB multiple
times for the same SHA-1.
This would explain partially why there was such a good
performance boost in the referenced commit above as we
implicitly lookup the same object multiple times.
Peff is really into getting this part faster, c.f.
https://public-inbox.org/git/20160914235547.h3n2otje2hec6u7k@xxxxxxxxxxxxxxxxxxxxx/
That looks interesting, but I question the probabilities for
my use case here. When walking the trees and files in a single
commit, I have no expectation that I'll see the same tree OID
twice, so the cache is not really useful and may just add
overhead. However, in a checkout or merge there is a high
expectation of visiting the same tree OID -- and most of the
time they are peers -- since commits tend to only change
isolated parts of the tree. (I'm not going to worry about the
case where someone moves an entire sub-tree to somewhere else
in the tree and violates my peer assertion.)
I did notice that we do 2 independent passes during checkout.
First to compare the old and new commits. Then to compare the
new with the worktree. So we touch each tree object 3 times.
My patch helps the first, but does nothing for the second.
Hopefully the cache is helping it (but I have not measured that).
TODO This change is a first attempt to test that by comparing
TODO the hashes of name[i] and name[i-i] and simply copying
TODO the tree-descriptor data. I was thinking of the n=2
TODO case here. We may want to extend this to the n=3 case.
================
On the Windows repo (500K trees, 3.1M files, 450MB index),
this reduced the overall time by 0.75 seconds when cycling
between 2 commits with a single file difference.
(avg) before: 22.699
(avg) after: 21.955
===============
So it shaves off 4% of the time needed. it doesn't sound like a
break through, but I guess these small patches add up. :)
Agreed, but on the Windows source repo, it can take 30 seconds to
do a "checkout -b" (without changing the actual HEAD commit).
That's just for the housekeeping of ensuring you get a clean
worktree. If I can knock off 5% here with minimal impact
and without changing any file formats, I'll take it.
And if I can just repeat that n times... :-)
for (i = 0; i < n; i++, dirmask >>= 1) {
- const unsigned char *sha1 = NULL;
- if (dirmask & 1)
- sha1 = names[i].oid->hash;
- buf[i] = fill_tree_descriptor(t+i, sha1);
+ if (i > 0 && (dirmask & 1) && names[i].oid && names[i-1].oid &&
+ !hashcmp(names[i].oid->hash, names[i-1].oid->hash)) {
Why do we need to check for dirmask & 1 here?
This ought to be covered by the hashcmp already IIUC.
So maybe we can pull out the
if (dirmask & 1)
sha1 = names[i].oid->hash;
out of the else when dropping that dirmask check?
I was wondering about the test in the else clause.
Just now I put a quick assert in there and it went off,
so I'm not going to change this.
Thanks
Jeff