Re: How much kernel memory is in 64-bit OS ?

[Kyle Moffett <mrmacman_g4@mac.com>]

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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