From 830ec0458c390f29c6c99e1ff7feab9e36368d12 Mon Sep 17 00:00:00 2001 From: John Stultz Date: Thu, 18 Mar 2010 14:47:30 -0700 Subject: time: Fix accumulation bug triggered by long delay. The logarithmic accumulation done in the timekeeping has some overflow protection that limits the max shift value. That means it will take more then shift loops to accumulate all of the cycles. This causes the shift decrement to underflow, which causes the loop to never exit. The simplest fix would be simply to do a: if (shift) shift--; However that is not optimal, as we know the cycle offset is larger then the interval << shift, the above would make shift drop to zero, then we would be spinning for quite awhile accumulating at interval chunks at a time. Instead, this patch only decreases shift if the offset is smaller then cycle_interval << shift. This makes sure we accumulate using the largest chunks possible without overflowing tick_length, and limits the number of iterations through the loop. This issue was found and reported by Sonic Zhang, who also tested the fix. Many thanks your explanation and testing! Reported-by: Sonic Zhang Signed-off-by: John Stultz Tested-by: Sonic Zhang LKML-Reference: <1268948850-5225-1-git-send-email-johnstul@us.ibm.com> Signed-off-by: Thomas Gleixner --- kernel/time/timekeeping.c | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) (limited to 'kernel/time/timekeeping.c') diff --git a/kernel/time/timekeeping.c b/kernel/time/timekeeping.c index 16736379a9ca..39f6177fafac 100644 --- a/kernel/time/timekeeping.c +++ b/kernel/time/timekeeping.c @@ -818,7 +818,8 @@ void update_wall_time(void) shift = min(shift, maxshift); while (offset >= timekeeper.cycle_interval) { offset = logarithmic_accumulation(offset, shift); - shift--; + if(offset < timekeeper.cycle_interval<