On Mon, Aug 3, 2009 at 11:16 PM, Michael Buesch<[email protected]> wrote:
> On Monday 03 August 2009 22:58:30 Larry Finger wrote:
>> Gábor states it the way the Broadcom routine is written. They have the
>> flags divided into 3 16-bit values - high, middle, and low. The values
>> are kept in arrays - one set is for the current band and the other is
>> for both bands. When the routine is entered, the appropriate quantity
>> is saved in a temporary, then the array value is maskset. Only when
>> the resulting value changes is the shared memory location updated. The
>> implication is that shared memory writes are expensive. Is that true?
>
> No. I think it has other reasons.
>
> --
> Greetings, Michael.
>

In that case, is the code correct:

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