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!