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

["Frank Schäfer" <fschaefer.oss@googlemail.com>]

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 ?

Regards,
Frank

> Regards,
> Mauro