Am 13.04.2013 19:46, schrieb Frank Schäfer: > Am 13.04.2013 19:04, schrieb Mauro Carvalho Chehab: >> Em Sat, 13 Apr 2013 17:33:28 +0200 >> Frank Schäfer <fschaefer.oss@xxxxxxxxxxxxxx> escreveu: >> >>> Am 13.04.2013 16:41, schrieb Mauro Carvalho Chehab: >>>> Em Sat, 13 Apr 2013 11:48:39 +0200 >>>> Frank Schäfer <fschaefer.oss@xxxxxxxxxxxxxx> 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@xxxxxxxxxxxxxx> >>>>> --- >>>>> 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. > I don't get it. > We always write to the GPIO register. That's why these functions are > called em28xx_write_* ;) > Whether the write operation is sane or not (e.g. because it modifies the > bit corresponding to an input line) is not subject of these functions. Hmm... that's actually not true for em28xx_write_regs(). The current/old code never writes the value to GPIO registers, it just saves it to the device struct. IMHO, this is plain wrong and yet antoher reason for applying this patch. ;) It just didn't cause any trouble (hopefully) because for the GPIO registers em28xx_write_reg_bits() is usually used instead (which works correctly). After checking the whole GPIO stuff again, I noticed a different potential problem: Register 0x04 seems to be a pure GPO register, so it is possible that reading the current value from this register doesn't work. The note in em28xx_write_regs() implies that noone has ever tested if it works correctly. Anyway, the current code reads register 0x04, too, to get the initial value for caching. ;) Regards, Frank > > > Frank > -- To unsubscribe from this list: send the line "unsubscribe linux-media" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
