Re: [PATCH 1/3] em28xx: give up GPIO register tracking/caching

[Mauro Carvalho Chehab <mchehab@redhat.com>]

Em Sat, 13 Apr 2013 17:33:28 +0200
Frank Schäfer <fschaefer.oss@googlemail.com> escreveu:

> Am 13.04.2013 16:41, schrieb Mauro Carvalho Chehab:
> > Em Sat, 13 Apr 2013 11:48:39 +0200
> > Frank Schäfer <fschaefer.oss@googlemail.com> escreveu:
> >
> >> The GPIO register tracking/caching code is partially broken, because newer
> >> devices provide more than one GPIO register and some of them are even using
> >> separate registers for read and write access.
> >> Making it work would be too complicated.
> >> It is also used nowhere and doesn't make sense in cases where input lines are
> >> connected to buttons etc.
> >>
> >> Signed-off-by: Frank Schäfer <fschaefer.oss@googlemail.com>
> >> ---
> >>  drivers/media/usb/em28xx/em28xx-cards.c |   12 ------------
> >>  drivers/media/usb/em28xx/em28xx-core.c  |   27 ++-------------------------
> >>  drivers/media/usb/em28xx/em28xx.h       |    6 ------
> >>  3 Dateien geändert, 2 Zeilen hinzugefügt(+), 43 Zeilen entfernt(-)
> > ...
> >
> >
> >> @@ -231,14 +215,7 @@ int em28xx_write_reg_bits(struct em28xx *dev, u16 reg, u8 val,
> >>  	int oldval;
> >>  	u8 newval;
> >>  
> >> -	/* Uses cache for gpo/gpio registers */
> >> -	if (reg == dev->reg_gpo_num)
> >> -		oldval = dev->reg_gpo;
> >> -	else if (reg == dev->reg_gpio_num)
> >> -		oldval = dev->reg_gpio;
> >> -	else
> >> -		oldval = em28xx_read_reg(dev, reg);
> >> -
> >> +	oldval = em28xx_read_reg(dev, reg);
> >>  	if (oldval < 0)
> >>  		return oldval;
> > That's plain wrong, as it will break GPIO input.
> >
> > With GPIO, you can write either 0 or 1 to a GPIO output port. So, your
> > code works for output ports.
> >
> > However, an input port requires an specific value (either 1 or 0 depending
> > on the GPIO circuitry). If the wrong value is written there, the input port
> > will stop working.
> >
> > So, you can't simply read a value from a GPIO input and write it. You need
> > to shadow the GPIO write values instead.
> 
> I don't understand what you mean.
> Why can I not read the value of a GPIO input and write it ?

Because, depending on the value you write, it can transform the input into an
output port.

If you don't understand why, I suggest you to take a look on how the GPIO
circuits are implemented. A very quick explanation could be find here:
	http://www.mcc-us.com/Open-collectorFAQ.htm

A more detailed one could be find here:

	http://www.coactionos.com/embedded-design/28-using-pull-ups-and-pull-downs.html


So, looking at the picture at http://www.coactionos.com/images/pullup.png and
assuming that a 0 means that the MOSFET gate is open, 1 means it is closed, 
for a pull-up GPIO input pin to work, driver needs to write "1" on it, so that
it will have VCC there.

This way, when MOSFEG goes to 1, the GPIO will be short-ciruited with GND, and
the driver will read a 0.

Note, however, that, if the driver writes a 0 to GPIO, no matter if the MOSFET
is opened or closed, it will read 0 every time.

Just the opposite logic applies for the pull-down logic.


-- 

Cheers,
Mauro