Re: Linux 2.4.13-ac4

Henning P. Schmiedehausen
2001-10-29T17:14:42Z

Vojtech Pavlik <vojtech@suse.cz> writes:

>On Mon, Oct 29, 2001 at 10:56:35AM +0000, Alan Cox wrote:
>> > bytes read from the 8254 get swapped. I've got some indirect evidence
>> > that this also could happen with the original i8254.
>>
>> Im hoping not. That would imply we interrupted someone half way through
>> reading the counter which means the locking is screwed up.

>Some old DOS assembly sources say that this sometimes happens without
>any interrupts at all - just that the chip is sometimes confused. It'd

Hm. If this is like the 8253 (ugh, way way back in the good old 8085
days I really wired and programmed such a bugger on my CP/M system...),
then the problem is, that the 16 bit counter is read in two 8 bit portions.
And if you do this:

Hi Lo

00000000 11111111 ----> read low byte: 11111111

Counter increments

00000001 00000000 ----> read high byte: 00000001

=> Counter is 00000001 11111111 which is wrong. :-)

Sames goes for decrement, I have forgotten in which direction this
bugger counts.

There was a trick when CPUs had much less MHz:

read low byte -> A
read high byte -> B
read low byte -> C

If A and C are the same, then the counter didn't change while reading:

byte a, b, c;

a = inb(lowbyte);

for(;;) {
b = inb(hibyte);
c = inb(lobyte);
if(c == a)
break;
a = c;
}

result = b<<8+a;

This assumes that you can't get _two_ changes in the high byte while
reading the ports in a row.

>be definitely worth printing the bad and good values when a problem is
>detected, so that we know what's happening.

This will happen all the time, so printing out is neither a good idea
nor is the read problem described above an error. It is just a quirk
in using an 8 bit chip in an 16 bit environment without being able to
"latch" the count.

Regards
Henning

--
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