linux-kernel - Re: [PATCH v2 1/1] x86/boot/compressed: Fix avoiding memmap in physical KASLR

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Message-ID: <6bd76106-339f-4204-a418-738b7ee545ab@intel.com>
Date: Thu, 23 Oct 2025 13:29:13 -0700
From: Dave Hansen <dave.hansen@...el.com>
To: Michal Clapinski <mclapinski@...gle.com>,
 Thomas Gleixner <tglx@...utronix.de>, Ingo Molnar <mingo@...hat.com>,
 Borislav Petkov <bp@...en8.de>, Dave Hansen <dave.hansen@...ux.intel.com>,
 Chris Li <chrisl@...nel.org>, x86@...nel.org
Cc: "H. Peter Anvin" <hpa@...or.com>, Ard Biesheuvel <ardb@...nel.org>,
 Dan Williams <dan.j.williams@...el.com>,
 Pasha Tatashin <pasha.tatashin@...een.com>, linux-kernel@...r.kernel.org
Subject: Re: [PATCH v2 1/1] x86/boot/compressed: Fix avoiding memmap in
 physical KASLR

On 10/22/25 16:37, Michal Clapinski wrote:
...
> But it would not disable physical KASLR for:
> memmap=1G!4G memmap=1G!5G memmap=1G!6G memmap=1G!7G memmap=1G!8G
> since the whole function would be called 5 times and the last `if`
> would never trigger.

I'm missing something about how this works.

The:

        static int i;

is static so should be keeping state across function calls. For the
purposes of checking 'i', why does it matter if the function is called
one time with 5 arguments or 5 times with 1? Doesn't 'i' end up at the
same value either way?