Hi Wolfgang,
On Fri, Apr 1, 2011 at 3:39 PM, Wolfgang Denk wd@denx.de wrote:
Dear Lei Wen,
In message AANLkTindN8kSqiq28H68Rs77Dc6Pf-4MSTibAq_2dRaV@mail.gmail.com you wrote:
This makes no sense to me. I have never seen any 9 bit registers in any processor I ever encountered in real life.
I don't mean that register is 9bit... I means that register, IER, is 32bit long, but 9-31th bit is reserved, and 0th to 8th bit is used... Maybe I don't say clearly... So byte access would only cover 0-7th bit, while 8th bit is not covered.
Also, in this case the serial controller is probably not NS16550 compatible, because AFAICT the NS16550 uses only 8 bit wide registers.
This is may be additional feature added. For another part except this one bit is all compatible with ns16550.
OK, so let's summarize the facts we found so far:
- Your hardware is NOT NS16550 compatible. It does not have the
typical 8 bit register interface, but provides additional bits that somehow control non-standard functionality.
- You say there is one additional bit (bit 9, i. e. 0x00000100) which
must remain set to 1 which appears to be the default setting after power-on reset.
- You say that the current implementation, which uses a writeb() call
(i. e. a byte write operation) to this register would not only affect bits 0...7, as expected, but also clear bit 9.
That is not my case. In my case, for writeb, it would affect only bits0-7, but leave bit 8 untouched. However, I need the bit 8 to be set to be 0, which is 1 at the power on.
This seems somewhat unlikely to me, so please confirm that I understood correctly.
- You say that writing a 32 bit value instead (i. e. using writel())
would work for you.
This seems also unlikely to me, as the driver just operates on 8 bit data, and will insert only 8 bit data into the word written, so you will write some data word like 0x000000XX - and in this case (and only in this case) the 9th bit would explicitly be set to 0.
Yes, that is what I want. The bit8 set to 0.
Best regards, Lei