Messages in this thread | | | Date | Sun, 01 Oct 2006 13:48:03 +1000 | From | Nick Piggin <> | Subject | Re: How is Code in do_sys_settimeofday() safe in case of SMP and Nest Kernel Path? |
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Christoph Lameter wrote: > On Sun, 1 Oct 2006, Dong Feng wrote: > > >>--- kernel/time.c.orig 2006-09-30 23:21:29.000000000 +0800 >>+++ kernel/time.c 2006-09-30 23:38:18.000000000 +0800 >>@@ -107,7 +107,16 @@ asmlinkage long sys_gettimeofday(struct >> return -EFAULT; >> } >> if (unlikely(tz != NULL)) { >>- if (copy_to_user(tz, &sys_tz, sizeof(sys_tz))) >>+ struct timezone ktz; >>+ unsigned long seq; >>+ >>+ do { >>+ seq = read_seqbegin(&xtime_lock); >>+ ktz.tz_minuteswest = sys_tz.tz_minuteswest; >>+ ktz.tz_dsttime = sys_tz.tz_dsttime; >>+ } while (unlikely(read_seqretry(&xtime_lock, seq))); >>+ >>+ if (copy_to_user(tz, &ktz, sizeof(ktz))) >> return -EFAULT; > > > I really hate adding overhead to gettimeofday() and we would have to take > the seqlock in all places when we reference tz. Maybe we can tolerate the > resulting race? > > If we assume word size transfers then we only have an issue on 32 bit > platforms. The result of the race would be that tz_minuteswest and > tz_dsttime disagree. So we may get daylight savings time wrong. > But then we are already changing the timezone and are potentially warping time. > gettimofday may be unstable anyways. So it may be okay to leave the race > in. Just add some comments explaining the situation.
It is in an unlikely path though. How many apps actually pass in a non NULL value for the timezone? Those that don't won't be affected. Even for those that do, it doesn't introduce any atomic ops or unpredictable branches, or cacheline pressure (because xtime lock is already touched by do_gettimeofday). IOW: I'm sure it would be unmeasurable.
OTOH, to be completely correct, it seems like the same xtime_lock read section should cover both the calculation of ktv, and that of ktz. So if it is going to be fixed at all, it should be done properly and looks like it needs to be a bit more intrusive (but no more expensive).
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