Re: [2.4.28-rc1] process stuck in release_task() call

[Willy Tarreau <willy@w.ods.org>]

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 ?

  - would it enhance performance a bit to put a bunch of 'unlikely()' in all
    the ifs which end in BUG(), especially inside the loop ?

Regards,
Willy