Re: clear_bit_unlock_is_negative_byte

[Matthew Wilcox <willy@infradead.org>]

On Fri, Jul 21, 2023 at 01:12:48PM +1200, Michael Schmitz wrote:
> Logging the bit nr. and value passed in:
> 
> nfhd8: found device with 20971440 blocks (512 bytes)
> cbu_inb memval mismatch: 2004 12005 2005 0
> cbu_inb memval mismatch: 2004 12005 2005 0
>  nfhd8: AHDI p1 p2
> cbu_inb memval mismatch: 36 10037 37 0
> cbu_inb memval mismatch: 36 10037 37 0

I'm not quite sure what values you're printing here?  And whether
they're hex or decimal.  Could you show me the printk string?

> The instruction you need is eori.b, and you'll have to increase the mem
> pointer by 3 bytes. With that change, I see no further mismatches until the
> return values begin to differ once disk access begins:

Ah thanks.  The perils of working from Motorola official docs and then
trying to use the GNU assembler ...

> sd 0:0:1:0: [sdb] Preferred minimum I/O size 512 bytes
> cbu_inb retval mismatch: 1 ff 2084 2084 2085 0
> rtc-generic rtc-generic: registered as rtc0
> cbu_inb retval mismatch: 1 ff 2094 2094 2095 0
> ...
> sd 0:0:0:0: [sda] Attached SCSI disk
> probe of 0:0:0:0 returned 0 after 58395182 usecs
> cbu_inb retval mismatch: 1 ff 2094 2094 2095 0
>  sdb: RDSK (512) sdb1 (DOS^G)(res 2 spb 2) sdb2 (SFS^B)(res 2 spb 1) sdb3
> (SFS^B)(res 2 spb 2) sdb4 ((res 2 spb 1)
> sdb: p4 size 18446744071971831216 extends beyond EOD, enabling native
> capacity
> cbu_inb retval mismatch: 1 ff 2084 2084 2085 0
> 
> (return value from old and new code, value of mem from old and new code,
> original value, bit nr).

OK, so the new code would set the byte to 0xff (that's how Sxx works).
But it's returning a bool, so that shouldn't matter.  2084/2084/2085
would make sense; one of the two functions has cleared the bottom bit.

> Bit 7 was already set before xor, and wasn't cleared. I suspect that's why
> the return value is no longer 1?

Ah, it's not supposed to be cleared.  The way this works is that bit 0
is the lock bit; if someone's waiting on the folio, they set bit 7.  If
bit 7 is set when we clear bit 0, we look on the wait queue.  If there's
nobody on the wait queue, we clear bit 7.

So ... I think you've fixed it!