On 11/20/19 12:02 AM, Tzung-Bi Shih wrote:
max98090_interrupt() and max98090_pll_work() run in 2 different threads.
There are 2 possible races:
Note: M98090_REG_DEVICE_STATUS = 0x01.
Note: ULK == 0, PLL is locked; ULK == 1, PLL is unlocked.
max98090_interrupt max98090_pll_work
----------------------------------------------
schedule max98090_pll_work
restart max98090 codec
receive ULK INT
assert ULK == 0
schedule max98090_pll_work (1).
In the case (1), the PLL is locked but max98090_interrupt unnecessarily
schedules another max98090_pll_work.
if you re-test that the PLL is already running, then you can exit the
work function immediately without redoing the sequence?
maybe also play with the masks so that the PLL unlock is masked in the
interrupt and unmasked after the PLL locks?
max98090_interrupt max98090_pll_work max98090 codec
----------------------------------------------------------------------
ULK = 1
receive ULK INT
read 0x01
ULK = 0 (clear on read)
schedule max98090_pll_work
restart max98090 codec
ULK = 1
receive ULK INT
read 0x01
ULK = 0 (clear on read)
read 0x1
assert ULK == 0 (2).
what are those 0x01 and 0x1? is the second a typo possibly?
In the case (2), both max98090_interrupt and max98090_pll_work read
the same clear-on-read register. max98090_pll_work would falsely
thought PLL is locked.
There are 2 possible options:
A. turn off ULK interrupt before scheduling max98090_pll_work; and turn
on again before exiting max98090_pll_work.
B. remove the second thread of execution.
Adopts option B which is more straightforward.
but has the side effect of possibly adding a 10ms delay in the interrupt
thread?
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