Messages in this thread |  | | Subject | Re: [PATCH 3/6] cpuidle: menu: Get rid of first_idx from menu_select() | From | Daniel Lezcano <> | Date | Fri, 5 Oct 2018 10:35:00 +0200 |
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On 04/10/2018 19:19, Rafael J. Wysocki wrote: > On Thu, Oct 4, 2018 at 4:51 PM Daniel Lezcano <daniel.lezcano@linaro.org> wrote: >> >> On Tue, Oct 02, 2018 at 11:44:06PM +0200, Rafael J. Wysocki wrote: >>> From: Rafael J. Wysocki <rafael.j.wysocki@intel.com> >>> >>> Rearrange the code in menu_select() so that the loop over idle states >>> always starts from 0 and get rid of the first_idx variable. >>> >>> While at it, add two empty lines to separate conditional statements >>> one another. >>> >>> No intentional behavior changes. >>> >>> Signed-off-by: Rafael J. Wysocki <rafael.j.wysocki@intel.com> >>> --- >> >> This code is becoming a bit complex to follow :/ >> >> May be I missed something, but it is not possible to enter the condition without >> idx != 0, no ? (I meant the condition if ((drv->states[idx].flags & >> FLAG_POLLING)) > > Not sure what you mean.
Yes, sorry let me clarify.
I meant the flag polling is always on state[0], so if the flag is set then idx == 0.
We have the conditions:
drv->states[idx].flags & CPUIDLE_FLAG_POLLING
If it is true then idx is zero.
Then comes the second condition:
i == idx + 1
because of the above, idx is zero, then it can become i == 1.
Then the variable assignation:
idx = i can be replaced by idx = 1
> We start with idx = -1, i = 0. If state[0] is enabled, idx becomes 0. > > If the target residency or exit latency of state[0] are already beyond > the limits, we just bail out and state[0] will be returned. > > If not, we go to i = 1, but idx is still 0. If the target residency > of state[1] is beyond predicted_us (which is the interesting case), we > check (drv->states[0].flags & FLAG_POLLING) and so on. > > Currently, the polling state must be state[0] (if used at all) anyway. > >> Reviewed-by: Daniel Lezcano <daniel.lezcano@linaro.org> >> > > Thanks! >
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