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Message-Id: <A01B0745-4715-452E-847E-28F28C6EE34F@mac.com>
Date: Mon, 18 Sep 2006 09:30:43 -0400
From: Kyle Moffett <mrmacman_g4@....com>
To: Jan Engelhardt <jengelh@...ux01.gwdg.de>
Cc: Rik van Riel <riel@...hat.com>,
yogeshwar sonawane <yogyas@...il.com>,
linux-kernel <linux-kernel@...r.kernel.org>
Subject: Re: How much kernel memory is in 64-bit OS ?
On Sep 18, 2006, at 08:34:25, Jan Engelhardt wrote:
>>>> Sure, people said that too when going from 16 bits to 32 bits,
>>>> but that was only a factor 2^16 difference. This time it's the
>>>> square of the previous difference.
>>>
>>> Not quite the square :)
>>
>> 2^32 is the square of 2^16 :)
>
> As mentioned above, "the square of the previos" [16 to 32]
> "difference". Call me nitpicky, but:
>
> (2^32 - 2^16)^2 != (2^64 - 2^32)
Well into the nitpicking territory... I think you missed where he
said "factor 2^16 difference". So these:
> Going from 32 to 64: (2^64 - 2^32)
> Going from 16 to 32: (2^32 - 2^16)
Should actually be these:
Going from 32 to 64: (2^64 / 2^32)
Going from 16 to 32: (2^32 / 2^16)
And:
(2^32 / 2^16)^2 = (2^64 / 2^32) => (2^16)^2 = (2^32)
Cheers,
Kyle Moffett
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