linux-kernel - Re: [PATCH WIP v1 07/20] mm/rmap_id: track if one ore multiple MMs map a partially-mappable folio

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Message-Id: <4FFED5A1-BF86-480D-8CB7-BECE5C413B44@gmail.com>
Date: Mon, 18 Dec 2023 16:34:42 +0200
From: Nadav Amit <nadav.amit@...il.com>
To: David Hildenbrand <david@...hat.com>
Cc: Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
 linux-mm <linux-mm@...ck.org>,
 Andrew Morton <akpm@...ux-foundation.org>,
 Linus Torvalds <torvalds@...ux-foundation.org>,
 Ryan Roberts <ryan.roberts@....com>,
 Matthew Wilcox <willy@...radead.org>,
 Hugh Dickins <hughd@...gle.com>,
 Yin Fengwei <fengwei.yin@...el.com>,
 Yang Shi <shy828301@...il.com>,
 Ying Huang <ying.huang@...el.com>,
 Zi Yan <ziy@...dia.com>,
 Peter Zijlstra <peterz@...radead.org>,
 Ingo Molnar <mingo@...hat.com>,
 Will Deacon <will@...nel.org>,
 Waiman Long <longman@...hat.com>,
 "Paul E. McKenney" <paulmck@...nel.org>
Subject: Re: [PATCH WIP v1 07/20] mm/rmap_id: track if one ore multiple MMs
 map a partially-mappable folio

> On Dec 18, 2023, at 4:04 PM, David Hildenbrand <david@...hat.com> wrote:
> 
> But adding 1 "0" bit is not sufficient for handling order-2 folios (P = 4), only for handling order-1 folios. So what the current approach does is the following (P = 4):
> 
> RMAP-ID |       | Subid |
> -----------------------------------
> 0       | 0000 | 0     | 0000 0000
> 1       | 0001 | 1     | 0000 0001
> 2       | 0010 | 5     | 0000 0101
> 3       | 0011 | 6     | 0000 0110
> 4       | 0100 | 25    | 0001 1001
> 5       | 0101 | 26    | 0001 1010
> 6       | 0110 | 30    | 0001 1110
> 7       | 0111 | 31    | 0001 1111
> 8       | 1000 | 125   | 0111 1101
> 9       | 1001 | 126   | 0111 1110
> 10      | 1010 | 130   | 1000 0010
> 11      | 1011 | 131   | 1000 0011
> 12      | 1100 | 150   | 1001 0110
> 13      | 1101 | 151   | 1001 0111
> 14      | 1110 | 155   | 1001 1011
> 15      | 1111 | 156   | 1001 1100

Yes, of course. Silly me. You want to take advantage of the counter not
saturating for orders K where K-1 is not a power of 2.

I get your point. Not sure whether it worth the complexity though…