Re: [RFC] on general object IDs again

[Pavel Emelyanov <xemul@parallels.com>]

On 01/11/2012 11:29 PM, KOSAKI Motohiro wrote:
>>>> Then, you only need to compare. not any other calculation. i.e. only
>>>> need id uniqueness.
>>>> And any resource are referenced from tasks. so, can you reuse pid for
>>>> this? example,
>>>> two taska share one mm.
>>>>
>>>> task-a(pid: 100)
>>>>               |-----------------mm
>>>> task-b(pid: 200)
>>>>
>>>>
>>>> gen_obj_id(task-b, GEN_OBJ_ID_VM) return 100. (youngest pid of referenced tasks)
>>>
>>> We can, but determining the youngest pid for an mm struct is O(N) algo.
>>> Having N tasks with N mm_structs getting the sharing picture becomes O(N^2).
>>
>> Yeah, exactly. If not the speed problem we would simply stick
>> with Andrew's proposal as two-id-are-the-same(pid1, pid2)
>> syscall.
> 
> Why O(N^2) is matter? Typical HPC system have mere a few hundred pids.
> so, O(N^2)
> is not slow. How do you mesure Andrew's proposal?
> 
> If you have 1000 pids and each syscall need 10usec,
> 
> 1000 * 1000 * 10 = 10,000,000usec = 10sec. But, important thing is, almost all
> processes don't share fs, mm and other structs. then, if we check
> reference count
> before task traversal, required time may reduce 1/10x - 1/100x.

This might work for mm_structs, although quite a lot apps now do have threads and
this mm->users check will be negative. But how about open files? Once we entered the
get-the-youngest-file-owner routine we need to take locks and with 1000 tasks the
overhead is not 1000 syscalls, but 1000 (syscalls + locks).

> 
>> But when we get a number of pids to dump we need the
>> resource affinity picture over them all.
> .
>