Daniel Shahaf wrote on Thu, May 26, 2011 at 00:36:48 +0300:
> I've not read the code, but would an array of 'svn_revnum_t' be
> a better representation?
> Specifically: an append-only array of [from_rev, to_rev] pairs, sorted
> by from_rev. That's less overhead (and we could take advantage of the
> sorting to store less data), at the cost of O(log(n)) lookup.
And the numbers... well, for a plain C array it would be 70% less than
in Greg's analysis of the hash case, since the additional 20 bytes per
mapping entry are gone. (and that's before I suggest to not store
[N+1, M+1] if [N,M] is already stored)
> Greg Stein wrote on Wed, May 25, 2011 at 17:21:44 -0400:
> > On Wed, May 25, 2011 at 16:08, C. Michael Pilato <cmpilato_at_collab.net> wrote:
> > > On 05/25/2011 04:05 PM, C. Michael Pilato wrote:
> > >> On 05/25/2011 03:49 PM, Greg Stein wrote:
> > >>> On Wed, May 25, 2011 at 15:33, <cmpilato_at_apache.org> wrote:
> > >>>> ...
> > >>>> + /* A mapping of svn_revnum_t * dump stream revisions to their
> > >>>> + corresponding svn_revnum_t * target repository revisions. */
> > >>>> + apr_hash_t *rev_map;
> > >>>
> > >>> How big can this grow? ie. what happens when there are several million
> > >>> revisions.
> > >>
> > >> It gets big. (This logic and approach are copied from 'svnadmin load',
> > >> which doesn't excuse it, but might explain it.)
> > >
> > > Actually, I don't really know for sure how big it gets. It's a mapping of
> > > of sizeof(svn_revnum_t) to sizeof(svn_revnum_t), plus all the hash
> > > internals. Anybody have any guesses?
> > struct apr_hash_entry_t is generally 20 bytes. Add in the two revnums
> > (4 bytes each), and you get 28 bytes for each *used* entry.
> > Now we also have to account for unused entries. APR has a pretty poor
> > hash table implementation. It allocates *upwards* to the nearest power
> > of two. So the internal size will grow like:
> > 1048576
> > 2097152
> > 4194304
> > One saving grace is that APR only grows when the entry count matches
> > the internal table size. It uses a "closed hash" algorithm with linked
> > lists at each bucket, so the actual load on the buckets is not
> > possible to compute. The hand-wave means that you can put in 4 million
> > mappings before it grows it up to 8 million buckets.
> > So... 4 million buckets (pointers) at 4 bytes each is 80 megabytes.
> > Each mapping will add another 28 bytes. So: 4 million mappings is
> > about 134 megabytes. But also recognize that *reaching* that point
> > will use and toss approx the same amount of memory. So about 260 meg
> > total.
> > On a 64-bit architecture, all these values are likely to be doubled.
> > Not a machine crusher, in retrospect. But not exactly a winner either.
> > Cheers,
> > -g
Received on 2011-05-25 23:42:24 CEST