netdev - Re: [PATCH 0/1] Rosebush, a new hash table

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Date: Sun, 25 Feb 2024 13:32:40 +0800
From: Herbert Xu <herbert@...dor.apana.org.au>
To: Matthew Wilcox <willy@...radead.org>
Cc: Kent Overstreet <kent.overstreet@...ux.dev>,
	David Laight <David.Laight@...lab.com>,
	"linux-kernel@...r.kernel.org" <linux-kernel@...r.kernel.org>,
	Thomas Graf <tgraf@...g.ch>,
	"netdev@...r.kernel.org" <netdev@...r.kernel.org>,
	"linux-fsdevel@...r.kernel.org" <linux-fsdevel@...r.kernel.org>,
	"maple-tree@...ts.infradead.org" <maple-tree@...ts.infradead.org>,
	"rcu@...r.kernel.org" <rcu@...r.kernel.org>
Subject: Re: [PATCH 0/1] Rosebush, a new hash table

On Sun, Feb 25, 2024 at 05:01:19AM +0000, Matthew Wilcox wrote:
> 
> Task a trivial example where you have four entries unevenly distributed
> between two buckets, three in one bucket and one in the other.  Now 3/4
> of your lookups hit in one bucket and 1/4 in the other bucket.
> Obviously it's not as pronounced if you have 1000 buckets with 1000
> entries randomly distributed between the buckets.  But that distribution
> is not nearly as even as you might expect:
> 
> $ ./distrib
> 0: 362
> 1: 371
> 2: 193
> 3: 57
> 4: 13
> 5: 4

Indeed, that's why rhashtable only triggers a forced rehash at
a chain length of 16 even though we expect the average chain length
to be just 1.

The theoretical worst-case value is expected to be O(lg n/lg lg n).
However, I think 16 was picked because it was sufficient even for a
hash table that filled all memory.  Of course if anyone can provide
some calculation showing that this is insufficient I'm happy to raise
the limit.

Cheers,
-- 
Email: Herbert Xu <herbert@...dor.apana.org.au>
Home Page: http://gondor.apana.org.au/~herbert/
PGP Key: http://gondor.apana.org.au/~herbert/pubkey.txt