Michal Hocko wrote:
> On Sun 22-11-15 21:13:22, Tetsuo Handa wrote:
> > I was updating kmallocwd in preparation for testing "[RFC 0/3] OOM detection
> > rework v2" patchset. I noticed an unexpected result with linux.git as of
> > 3ad5d7e06a96 .
> >
> > The problem is that an OOM victim arrives at do_exit() with TIF_MEMDIE flag
> > set but without pending SIGKILL. Is this correct behavior?
>
> Have a look at out_of_memory where we do:
> /*
> * If current has a pending SIGKILL or is exiting, then automatically
> * select it. The goal is to allow it to allocate so that it may
> * quickly exit and free its memory.
> *
> * But don't select if current has already released its mm and cleared
> * TIF_MEMDIE flag at exit_mm(), otherwise an OOM livelock may occur.
> */
> if (current->mm &&
> (fatal_signal_pending(current) || task_will_free_mem(current))) {
> mark_oom_victim(current);
> return true;
> }
>
> So if the current was exiting already we are not killing it, we just give it
> access to memory reserves to expedite the exit. We do the same thing for the
> memcg case.
The result is the same even if I do
- BUG_ON(test_thread_flag(TIF_MEMDIE) && !fatal_signal_pending(current));
+ BUG_ON(test_thread_flag(TIF_MEMDIE) && !fatal_signal_pending(current) && !task_will_free_mem(current));
. I think that task_will_free_mem() is always false because this BUG_ON()
is located before "exit_signals(tsk); /* sets PF_EXITING */" line.
>
> Why would that be an issue in the first place?
The real problem I care is TIF_MEMDIE livelock.
MemAlloc: oom-tester4(11040) uninterruptible dying victim
MemAlloc: oom-tester4(11045) gfp=0x242014a order=0 delay=10000 dying
I'm not talking about TIF_MEMDIE livelock in this thread. I'm just worrying
that below output (which is caused by an OOM victim arriving at do_exit()
with TIF_MEMDIE flag set but without pending SIGKILL) is a foretaste of
unnoticed problem.
MemAlloc: oom-tester4(11520) uninterruptible victim
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