Re: [2.4.28-rc1] process stuck in release_task() call
From: Marcelo Tosatti
Date: Thu Nov 11 2004 - 06:37:23 EST
On Thu, Nov 11, 2004 at 09:33:12AM +0100, Willy Tarreau wrote:
> Hi Marcelo,
> > > >>EIP; c012073d <release_task+1fd/230> <=====
> > > c0120540 <release_task>:
> > > c0120540: 55 push %ebp
> > > ....
> > > c0120736: 89 d8 mov %ebx,%eax
> > > c0120738: e8 73 dd 01 00 call c013e4b0 <free_pages> <= here
> > is this release_task+1fd? Can you send me the full disassemble of release_task?
> Yes it is because the next instruction after call will be at c0120738+5 =
> c012073d = release_task+1fd. (the return address on the stack is the
> address of the next instruction after the call).
> > It can't be blocked here, its a "call" instruction.
> Seems rather strange indeed ! Perhaps this is not the disassembled function
> of the *running* kernel ? it would be good to disassemble vmlinux and ensure
> that it is exactly the one currently running. I too have already lost lots
> of time searching a wrong bug because I disassembled the wrong kernel, so
> I'm certain it can happen even when we're very careful :-(
> > free_pages can't block either. Odd.
> Marcelo, I have two questions for my own understanding :
> - free_pages does spin_lock(&zone->lock) around the while() loop.
> Considering that someone else could hold the lock (bug, etc...), it
> could block here. But my feeling is that if such a lock were kept held,
> the system would be totally frozen because everything which would want
> to free memory would get stuck (even a process exit). Am I right ?
Right, the system will be totally frozen spinning on the lock.
> - would it enhance performance a bit to put a bunch of 'unlikely()' in all
> the ifs which end in BUG(), especially inside the loop ?
Yes, it should generate better code.
Try it and see how the generated code differs from the original without unlikely.
I'm not aware of the internals of unlikely however, so I can't
explain how it works in details... the GCC documentation
should do it. :)
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