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SubjectRe: transmeta cpu code question
Quoting Ben Collins <bcollins@debian.org>:

> On Thu, Nov 20, 2003 at 03:02:18AM +0100, Nico Schottelius wrote:
> > Hello!
> >
> > What does this do:
> >
> > printk(KERN_INFO "CPU: Processor revision %u.%u.%u.%u,
> > %u MHz\n",
> > (cpu_rev >> 24) & 0xff,
> > (cpu_rev >> 16) & 0xff,
> > (cpu_rev >> 8) & 0xff,
> > cpu_rev & 0xff,
> > cpu_freq);
> >
> > (from arch/i386/kernel/cpu/transmeta.c)
> >
> > Does not & 0xff make no sense? 0 & 1 makes 0, 1 & 1 makes 1,
> > no changes.
> >
> > And I don't understand why we do this for 8bit and shifting the
> > cpu_rev...
>
> You are a bit confused. The cpu_rev is a 4 byte value, each byte is a
> decimal of the revision.
>
> And (0 & 1) makes 1, not 0. That's an AND, not an OR.

Last I checked, 0 AND anything is 0. Think of it this way:

A B A & B
------------
T F F
T T T
F T F
F F F

So, anything ANDed with F is false.

-sandalle

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eric@sandall.us | http://www.sourcemage.org/
http://eric.sandall.us/ | SysAdmin @ Inst. Shock Physics @ WSU
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