On Thu, 24 May 2018, Linus Torvalds wrote:

This doesn't seem to be taking 'param->min_size' into account.

It was in that rounded_hashtable_size() does, however, after more
thought I think we can do better by taking it much more into account.

I'm not sure that matters, but right now, if you have nelem_hint set and a
min_size, the min_size is honored (if you have just min_size it's already
ignored because the rhashtable always starts with HASH_DEFAULT_SIZE). So I
could imagine that somebody uses it to guarantee something. The docs say
that "min_size" is the minimum size for *shrinking* not for initializing,
so I guess it's debatable.

Also, wouldn't it make sense to make this all be a while loop? Or are you
just depending on the knowledge that HASH_DEFAULT_SIZE / 2 is already
guaranteed to be so small that there's no point? A comment to that effect
would be good, perhaps.

Yes, this is why I didn't loop. With the default size of 64 buckets, we
allocate 640 + 128 = 768 bytes for the tbl and the lock array, respectively.
By halving this, upon retrying, I was relying on it being to "small to fail".

However, after how about the resize being based on HASH_MIN_SIZE instead of
HASH_DEFAULT_SIZE? That way the initial table would be a _lot_ smaller and
aid the allocator that much more; which is why we're here in the first place.
Any performance costs of collisions would be completely unimportant in this
scenario.

Considering that some users set p.min_size to be rather large-ish (up to 1024
buckets afaict), we'd need the following:

	size = min(ht->p.min_size, HASH_MIN_SIZE);

Which takes into account the min_size = max(ht->p.min_size, HASH_MIN_SIZE)
which came before, thus p.min_size == 0 is already taken into account.

Thanks,
Davidlohr