[Openmcl-devel] Hash Table anomaly -- hash-table-size decreases - wondering how this can happen
rwiker at gmail.com
Sat Jan 3 00:26:45 PST 2015
My understanding (as far as it goes) is that hash-table-size is just a hint to the runtime of the expected number of keys in the hash table; it is used when creating a hash table and can later be used to create other hash tables of similar size. hash-table-count, on the other hand, is the actual, current number of keys in the hash table. I would expect hash-table-size to increase as the number of keys in the table grows past the initial size, and it would also make sense for hash-table-size to decrease if the hash table is rehashed as part of gc.
> On 03 Jan 2015, at 08:47 , Glenn Iba <giba at alum.mit.edu> wrote:
> Call for help!
> I'm doing some large searches in CCL, and have been using large hash-tables,
> but I"m perplexed that the hash-table-size is getting mysteriously decreased.
> Can anyone explain how this is possible?
> My speculation is that I'm exhausting the heap (though I don't get any notification of this),
> and that CCL is trying to create more heap space by shrinking my large hash-table.
> Does this sound like it could be possible? I'd prefer to get a notification that I'm out of space.
> Is there any way to control this?
> I'm running CCL 1.10 on a Mac with OS X Yosemite (10.10.1), with 8GB RAM.
> I'm creating a single large hash-table with
> (make-hash-table :test #'equalp :size 100000000) ;; 100,000,000
> I'm storing positions of my search space (each represented by a byte-vector of 16 unsigned-bytes)
> in this hash-table
> For each generation, I collect all the positions in the hash-table (to avoid duplicates).
> I then write the generation out to a file, and do CLRHASH so I can reuse the hash-table.
> My searches reach a point (as the generation size grows) when the hash-table-size
> decreases dramatically (from 100,000,000 to 12,396,373) -- how is this possible?
> I'd be happy to supply code, detailed traces, and whatever other info I can
> to anyone who'd be willing to help me figure this out.
> Thanks in advance!
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