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sched: Eliminate the ts->idle_lastupdate field
Now that the only user of ts->idle_lastupdate is update_ts_time_stats(), the entire field can be eliminated. In update_ts_time_stats(), idle_lastupdate is first set to "now", and a few lines later, the only user is an if() statement that assigns a variable either to "now" or to ts->idle_lastupdate, which has the value of "now" at that point. Signed-off-by: Arjan van de Ven <[email protected]> Signed-off-by: Andrew Morton <[email protected]> Reviewed-by: Rik van Riel <[email protected]> Acked-by: Peter Zijlstra <[email protected]> Cc: [email protected] LKML-Reference: <[email protected]> Signed-off-by: Ingo Molnar <[email protected]>
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@@ -60,7 +60,6 @@ struct tick_sched {
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ktime_t idle_waketime;
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ktime_t idle_exittime;
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ktime_t idle_sleeptime;
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ktime_t idle_lastupdate;
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ktime_t sleep_length;
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unsigned long last_jiffies;
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unsigned long next_jiffies;
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@@ -158,16 +158,13 @@ update_ts_time_stats(struct tick_sched *ts, ktime_t now, u64 *last_update_time)
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{
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ktime_t delta;
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ts->idle_lastupdate = now;
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if (ts->idle_active) {
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delta = ktime_sub(now, ts->idle_entrytime);
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ts->idle_sleeptime = ktime_add(ts->idle_sleeptime, delta);
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ts->idle_entrytime = now;
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}
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if (ts->idle_active && last_update_time)
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*last_update_time = ktime_to_us(ts->idle_lastupdate);
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else
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if (last_update_time)
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*last_update_time = ktime_to_us(now);
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}
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