Re: transmeta cpu code question

[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.

Think about it this way. If cpu_rev == 0x01040801, then this would
produce:

(0x01040801 >> 24 & 0xff) -> (0x01 & 0xff) ->     0x01

(0x01040801 >> 16 & 0xff) -> (0x0104 & 0xff) ->   0x04

(0x01040801 >> 8  & 0xff) -> (0x010408 & 0xff) -> 0x08

(0x01040801 & 0xff)                            -> 0x01


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