On Mon, 17 Sep 2007 19:32:11 -0700 (PDT)
Trent Piepho <xyzzy@xxxxxxxxxxxxx> wrote:
>
> The polling loop to check for ACI to go down was more convoluted than it
> needed to be. New loop should be more efficient and it is a lot simpler. The
> old loop checked for a timeout before checking for ACI down, which could
> result in an erroneous timeout. It's only a failure if the timeout expires
> _and_ ACI is still high. There is nothing wrong with the timeout expiring
> while the task is sleeping if ACI went low.
>
This was already fixed by Rene's patch. You only simplified it.
> A polling loop to check for the device to leaving INIT mode is removed. The
> device must have already left init for the previous ACI loop to have finished.
>
> Signed-off-by: Trent Piepho <xyzzy@xxxxxxxxxxxxx>
>
> diff -r ec96283a273f isa/ad1848/ad1848_lib.c
> --- a/isa/ad1848/ad1848_lib.c Mon Sep 17 19:20:48 2007 -0700
> +++ b/isa/ad1848/ad1848_lib.c Mon Sep 17 19:22:36 2007 -0700
> @@ -203,9 +203,8 @@ static void snd_ad1848_mce_up(struct snd
>
> static void snd_ad1848_mce_down(struct snd_ad1848 *chip)
> {
> - unsigned long flags;
> - int timeout;
> - unsigned long end_time;
> + unsigned long flags, timeout;
> + int regsel;
>
> spin_lock_irqsave(&chip->reg_lock, flags);
> for (timeout = 5; timeout > 0; timeout--)
> @@ -222,50 +221,34 @@ static void snd_ad1848_mce_down(struct s
> #endif
>
> chip->mce_bit &= ~AD1848_MCE;
> - timeout = inb(AD1848P(chip, REGSEL));
> - outb(chip->mce_bit | (timeout & 0x1f), AD1848P(chip, REGSEL));
> - if (timeout == 0x80)
> + regsel = inb(AD1848P(chip, REGSEL));
> + outb(chip->mce_bit | (regsel & 0x1f), AD1848P(chip, REGSEL));
> + if (regsel == 0x80)
> snd_printk(KERN_WARNING "mce_down [0x%lx]: serious init problem - codec still busy\n", chip->port);
> - if ((timeout & AD1848_MCE) == 0) {
> + if ((regsel & AD1848_MCE) == 0) {
> spin_unlock_irqrestore(&chip->reg_lock, flags);
> return;
> }
>
> /*
> - * Wait for (possible -- during init auto-calibration may not be set)
> - * calibration process to start. Needs upto 5 sample periods on AD1848
> - * which at the slowest possible rate of 5.5125 kHz means 907 us.
> + * Wait for auto-calibration (AC) process to finish, i.e. ACI to go low.
> + * It may take up to 5 sample periods (at most 907 us @ 5.5125 kHz) for
> + * the process to _start_, so it is important to wait at least that long
> + * before checking. Otherwise we might think AC has finished when it
> + * has in fact not begun. It should take 128 (no AC) or 384 (AC) cycles
> + * for ACI to drop. This gives a wait of at most 70 ms with a more
> + * typical value of 3-9 ms.
> */
> - spin_unlock_irqrestore(&chip->reg_lock, flags);
> - msleep(1);
> - spin_lock_irqsave(&chip->reg_lock, flags);
> -
> - snd_printdd("(2) jiffies = %lu\n", jiffies);
> -
> - end_time = jiffies + msecs_to_jiffies(250);
> - while (snd_ad1848_in(chip, AD1848_TEST_INIT) & AD1848_CALIB_IN_PROGRESS) {
> + timeout = jiffies + msecs_to_jiffies(250);
> + do {
> spin_unlock_irqrestore(&chip->reg_lock, flags);
> - if (time_after(jiffies, end_time)) {
> - snd_printk(KERN_ERR "mce_down - auto calibration time out (2)\n");
> - return;
> - }
> msleep(1);
> spin_lock_irqsave(&chip->reg_lock, flags);
> - }
> -
> - snd_printdd("(3) jiffies = %lu\n", jiffies);
> -
> - end_time = jiffies + msecs_to_jiffies(100);
> - while (inb(AD1848P(chip, REGSEL)) & AD1848_INIT) {
> - spin_unlock_irqrestore(&chip->reg_lock, flags);
> - if (time_after(jiffies, end_time)) {
> - snd_printk(KERN_ERR "mce_down - auto calibration time out (3)\n");
> - return;
> - }
> - msleep(1);
> - spin_lock_irqsave(&chip->reg_lock, flags);
> - }
> - spin_unlock_irqrestore(&chip->reg_lock, flags);
> + regsel = snd_ad1848_in(chip, AD1848_TEST_INIT);
> + } while ((regsel & AD1848_CALIB_IN_PROGRESS) && time_before(jiffies, timeout));
Break this long line. You may calculate "regsel & AD1848_CALIB_IN_PROGRESS"
inside the loop and use it in the condition outside the loop too.
Regards,
Krzysztof
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