Re: TTY: tty_port questions

Richard Weinberger
2012-03-10T23:21:15Z

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Am 10.03.2012 23:51, schrieb Jiri Slaby:
> On 03/10/2012 11:26 PM, Richard Weinberger wrote:
>> Hi!
>
>> While moving UML's console driver to tty_port some strange things
>> happened. So, I have a few questions. :-)
>
>> The original driver did not implement tty_operations->hangup(). If
>> I implement it and call tty_port_hangup(), as Alan suggested, the
>> login fails on all TTYs except tty0. It fails because the opened
>> TTY returns EIO upon read()/write() after /bin/login called
>> vhangup().
>
>> The call chain is: vhangup() -> tty_vhangup_self() -> tty_vhangup()
>> -> __tty_hangup()
>
>> Within __tty_hangup() something happens that I don't fully
>> understand:
>
>> if (cons_filp) { if (tty->ops->close) for (n = 0; n < closecount;
>> n++) tty->ops->close(tty, cons_filp); } else if (tty->ops->hangup)
>> (tty->ops->hangup)(tty);
>
>> Login on tty0 works because cons_filp is not NULL and
>> tty->ops->close() is called. On the other hand login fails on every
>> other TTY because cons_filp remains NULL and the TTY hangs up.
>
>> Is there something missing in my hangup function?
>
>> If I omit tty_operations->hangup() and leave it, like the old
>> driver, NULL non-tty0 TTYs cannot be opened. (getty terminates
>> immediately because it cannot open any TTY.) open() retuns -EIO
>> because the TTY_CLOSING is set in tty->flags.
>
>> How can this be?
>
> Hmm, it looks like some process is sitting on a TTY which was hung.
> And the system offers this hung TTY to others on further opens. Could
> you check that there is no process with open TTY after the vhangup?
>

"lsof | grep tty" does not show anything else than tty0. :-\

Thanks,
//richard

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