On Mon, 28 Aug 2006, Michael Poole wrote:
> linux-os \(Dick Johnson\) writes:
>
>> On Mon, 28 Aug 2006, Alan Cox wrote:
>>
>>> Ar Llu, 2006-08-28 am 08:17 -0400, ysgrifennodd linux-os (Dick Johnson):
>>>> On Sat, 26 Aug 2006 [email protected] wrote:
>>>>
>>>>>> Or we could just add a standardised extra set of speed ioctls, but then
>>>>>> we need to decide what occurs if I set the speed and then issue a
>>>>>> termios call - does it override or not.
>>>>>
>>>>> Actually, we're not QUITE out of bits. CBAUDEX | B0 is not taken.
>>>>
>>>> B0 is not a bit (there are no bits in 0). It won't work.
>>>
>>> Well that is how it is implemented and everyone else seems happy. If it
>>> violates your personal laws of physics you'll just have to cope.
>>
>> It has nothing to do with 'personal laws of physics'. On all recent
>> implementations, B0 is 0, i.e., the absence of any bits set. Therefore,
>> there is no observable difference between CBAUDEX and CBAUDEX | B0,
>> as shown above. Therefore, as I stated, it won't work.
>
> What baud rate does your system define CBAUDEX | B0 to be? On my
B0 is 0 (zero), no bits. If you are trying to play semantic games and
claim B0 is 1, i.e., bit 0, then it would not be written as B0, it
would be written as B(0) or B:0. B0 is defined to be the baud-rate
used to hang up the modem. It is zero in all bits on most all
implementations including my Sun. On most recent Linux distributions,
CBAUDEX is (octal) 0010000. Since B0 is zero, ORing it into CBAUDEX
does nothing.
> AMD64 machine, both the x86-64 and i386 asm/termbits.h files skip
> CBAUDEX -- B38400 is 0000017 and B57600 is 0010001 (CBAUDEX | B50).
> The headers do not define any baud rate between those two, either by
> rate or by c_cflag value.
>
> Michael Poole
>
Cheers,
Dick Johnson
Penguin : Linux version 2.6.16.24 on an i686 machine (5592.62 BogoMips).
New book: http://www.AbominableFirebug.com/
_
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