Re: [PATCH -mm v4 6/9] atmel_serial: Split the interrupt handler

[michael]

Hi,
> On Wed, 30 Jan 2008 11:29:59 +0100
> michael <trimarchi@xxxxxxxxxxxxxxxx> wrote:
>   
> > > Now, _that_ is strange. I can't see anything that needs protection
> > > across that call; in fact, I think we can lock a lot less than what we
> > > currently do.
> > >
> > >   
> > >       
> > I explain it bad:
> > - with spin_lock the system seems, there is no problem with Valuntary 
> > Preeption and
> > Preemptible Kernel
> > - with full preemption  it runs but the serial line can't be used for 
> > receiving at
> > high bit rate (using lrz)
> >     
>
> ...but if you drop the spinlock across the call to
> tty_flip_buffer_push, you get an Oops?
>
> Could you post the Oops?
>
>   
So this code

/*
* Drop the lock here since it might end up calling
* uart_start(), which takes the lock.
*/
spin_unlock(&port->lock);

tty_flip_buffer_push(port->info->tty);

spin_lock(&port->lock);

Works with:
CONFIG_PREEMPT_RT=y
CONFIG_PREEMPT=y
CONFIG_PREEMPT_SOFTIRQS=y
CONFIG_PREEMPT_HARDIRQS=y
CONFIG_PREEMPT_BKL=y

but crash with:
# CONFIG_PREEMPT_NONE is not set
CONFIG_PREEMPT_VOLUNTARY=y
# CONFIG_PREEMPT_DESKTOP is not set
# CONFIG_PREEMPT_RT is not set
CONFIG_PREEMPT_SOFTIRQS=y
# CONFIG_PREEMPT_HARDIRQS is not set
# CONFIG_PREEMPT_BKL is not set
CONFIG_CLASSIC_RCU=y

Seems to work in the last config if I comment the spin_lock & 
spin_unlock call.
/*
* Drop the lock here since it might end up calling
* uart_start(), which takes the lock.

spin_unlock(&port->lock); */

tty_flip_buffer_push(port->info->tty);

/* spin_lock(&port->lock); */

It is not readable because I can't compile it with debugging information 
(poor memory system)
> > > > Complete Preemption (Real-Time) ok but the serials is just unusable due
> > > > to too many overruns (just using lrz)
> > > >     
> > > >         
> > > Is it worse than before? IIRC Remy mentioned something about
> > > IRQF_NODELAY being the reason for moving all this code to softirq
> > > context in the first place; does the interrupt handler run in hardirq
> > > context?
> > >
> > >   
> > >       
> > In the complete preemption yes.
> >     
>
> Which question did you answer "yes" to? That it's worse than before or
> that the interrupt handler runs in hardirq context (i.e. IRQF_NODELAY)?
>
>   
The interrupt handler run in IRQF_NODELAY context.
> > > I think you're right. Can you change it and see if it helps?
> > >
> > >   
> > >       
> > I just change it because I have corruption on receiving buffer. All
> > my test are done with this fix
> >     
>
> Ok.
>
>   
> > > I guess I didn't test it thoroughly enough with DMA
> > > disabled...slub_debug ought to catch such things, but not until we
> > > receive enough data to actually overflow the buffer.
> > >
> > >   
> > >       
> > I just test it I don't have
> > buffer overflow.
> >     
>
> No, I'd expect your allocation fix to take care of that. Or did you by
> any chance test without the fix and with slub_debug enabled?
>
>   
I just meant that the buffer never fills up to its size.
> > I protect with a spinlock the access to the register when we sending
> > from the tasklet. It is correct?
> >     
>
> I have no idea. Could you post some more specifics about what you
> modified, for example a diff?
>
>   

...
/* The interrupt handler does not take the lock */
spin_lock_irqsave(&port->lock, flags);
atmel_tx_chars(port);
spin_unlock_irqrestore(&port->lock, flags);

spin_lock(&port->lock);
...

The atmel_tx_chars using the serial device registers like the interrupt 
routine
and so I think that it is possible to have interference during send 
operation.

> Most of the tasklet is already protected by the spinlock, so you must
> be careful to avoid any lock recursion.
>
> Haavard
>
>   
Regards Michael
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