Re: [PATCH 1 of 5] rombios/keyboard: Don't needlessly poll the status register

[Andrew Cooper <andrew.cooper3@citrix.com>]

On 27/11/12 16:33, Tim Deegan wrote:
> At 16:41 +0000 on 26 Nov (1353948092), Andrew Cooper wrote:
>> Repeated polling of the status register is not going to change its value, so
>> don't needlessly take 8192 traps to Qemu when 1 will do.
> AFAICS the purpose of this loop is to handle the case where someone's
> actually typing at the time.  The 0x2000 is intended to make us wait
> long enough for the next keypress or autorepeat.
>
> Reducing the loop to _2_ outb(0x80)s seems strange -- is there some case
> in qemu where another character will be presented after 2 delays but not
> immediately?

The reduction to 2 instead of 1 is because the while loop uses a
pre-decrement on max.

The patch was chosen at the time to have minimal change to the code.

~Andrew

>
> If not, can we dispense with 'max' altogether and just have 
>     while ( (inb(0x64) & 0x01) ) { inb(0x60); outb(0x80, 0x00); }
> or even
>     while ( (inb(0x64) & 0x01) ) inb(0x60);
> ?
>
> Tim.
>
>> Signed-off-by: Andrew Cooper <andrew.cooper3@citrix.com>
>>
>> diff -r 0049de3827bc -r 1728fb789940 tools/firmware/rombios/rombios.c
>> --- a/tools/firmware/rombios/rombios.c
>> +++ b/tools/firmware/rombios/rombios.c
>> @@ -1805,12 +1805,12 @@ keyboard_init()
>>      while ( (inb(0x64) & 0x02) && (--max>0)) outb(0x80, 0x00);
>>  
>>      /* flush incoming keys */
>> -    max=0x2000;
>> +    max=2;
>>      while (--max > 0) {
>>          outb(0x80, 0x00);
>>          if (inb(0x64) & 0x01) {
>>              inb(0x60);
>> -            max = 0x2000;
>> +            max = 2;
>>              }
>>          }
>>  
>>
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-- 
Andrew Cooper - Dom0 Kernel Engineer, Citrix XenServer
T: +44 (0)1223 225 900, http://www.citrix.com