Messages in this thread | | | Date | Fri, 16 Mar 2007 14:58:57 +0300 | From | Pavel Emelianov <> | Subject | Re: [RFC] kernel/pid.c pid allocation wierdness |
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Eric Dumazet wrote: > On Friday 16 March 2007 11:57, Pavel Emelianov wrote: >> Oleg Nesterov wrote: >>> On 03/14, Eric W. Biederman wrote: >>>> Pavel Emelianov <xemul@sw.ru> writes: >>>>> Hi. >>>>> >>>>> I'm looking at how alloc_pid() works and can't understand >>>>> one (simple/stupid) thing. >>>>> >>>>> It first kmem_cache_alloc()-s a strct pid, then calls >>>>> alloc_pidmap() and at the end it taks a global pidmap_lock() >>>>> to add new pid to hash. >>> We need some global lock. pidmap_lock is already here, and it is >>> only used to protect pidmap->page allocation. Iow, it is almost >>> unused. So it was very natural to re-use it while implementing >>> pidrefs. >>> >>>>> The question is - why does alloc_pidmap() use at least >>>>> two atomic ops and potentially loop to find a zero bit >>>>> in pidmap? Why not call alloc_pidmap() under pidmap_lock >>>>> and find zero pid in pidmap w/o any loops and atomics? >>> Currently we search for zero bit lockless, why do you want >>> to do it under spin_lock ? >> Search isn't lockless. Look: >> >> while (1) { >> if (!test_and_set_bit(...)) { >> atomic_dec(&nr_free); >> return pid; >> } >> ... >> } >> >> we use two atomic operations to find and set a bit in a map. > > The finding of the zero bit is done without lock. (Search/lookup) > > Then , the reservation of the found bit (test_and_set_bit) is done, and > decrement of nr_free. It may fail because the search was done lockless.
:\ I do understand how this algorithm works. What I don't understand is why it is done so, if we take a global lock anyway.
> Finding a zero bit in a 4096 bytes array may consume about 6000 cycles on > modern hardware. Much more on SMP/NUMA machines, or on machines where > PAGE_SIZE is 64K instead of 4K :) > > You don't want to hold pidmad_lock for so long period.
OK, thanks. That's explanations looks good.
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