linux-kernel - Re: [PATCH] x86/mce: Dynamically size space for machine check records

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Message-ID: <3be03f9e-53cc-4b40-9933-5bf59db84477@intel.com>
Date: Thu, 29 Feb 2024 15:56:36 -0800
From: Sohil Mehta <sohil.mehta@...el.com>
To: Tony Luck <tony.luck@...el.com>
CC: Borislav Petkov <bp@...en8.de>, "Naik, Avadhut" <avadnaik@....com>,
	"x86@...nel.org" <x86@...nel.org>, "linux-edac@...r.kernel.org"
	<linux-edac@...r.kernel.org>, "linux-kernel@...r.kernel.org"
	<linux-kernel@...r.kernel.org>, "yazen.ghannam@....com"
	<yazen.ghannam@....com>, Avadhut Naik <avadhut.naik@....com>
Subject: Re: [PATCH] x86/mce: Dynamically size space for machine check records

On 2/29/2024 9:21 AM, Tony Luck wrote:

> Looking at this specific case, sizeof(struct mce_evt_llist) is 136. So
> the original version of this code picks order 7 to allocate in 128 byte
> units. But this means that every allocation of a mce_evt_llist will take
> two 128-byte blocks.
> 
> Net result is that the comment at the top of arch/x86/kernel/cpu/mce/genpool.c
> that two pages are enough for ~80 records was wrong when written. At
> that point struct mce_evt_llist was below 128, so order was 6, and each
> allocation took two blocks. So two pages = 8192 bytes divided by (2 * 64)
> results in 64 possible allocations.
> 

Thanks for the explanation. The part that got me is that I somehow
expected ilog2() to round-up and not round-down.

> But over time Intel and AMD added to the structure. So the current math
> comes out at just 32 allocations before the pool is out of space.
> 
> Yazen provided the right answer for this. Change to use order_base_2()
> 

Yes, I agree. order_base_2() is better than doing ilog2(struct size) +
1. In the rare scenario of the size exactly being a power of 2 we don't
need to add the +1.