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Date: Wed, 11 Jan 2012 23:36:16 +0400
From: Pavel Emelyanov <xemul@parallels.com>
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To: KOSAKI Motohiro <kosaki.motohiro@gmail.com>
CC: Cyrill Gorcunov <gorcunov@gmail.com>, LKML <linux-kernel@vger.kernel.org>,
        Andrew Morton <akpm@linux-foundation.org>,
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        Glauber Costa <glommer@parallels.com>,
        Andi Kleen <andi@firstfloor.org>, Matt Helsley <matthltc@us.ibm.com>,
        Pekka Enberg <penberg@kernel.org>,
        Eric Dumazet <eric.dumazet@gmail.com>,
        Vasiliy Kulikov <segoon@openwall.com>,
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        "Eric W. Biederman" <ebiederm@xmission.com>,
        Andrey Vagin <avagin@openvz.org>
Subject: Re: [RFC] on general object IDs again
References: <20120111161939.GI8752@moon> <CAHGf_=r1vrZbCTYNAp7e+Hbbxh2BnDXXrdsNckSkQVVrXQ=9KQ@mail.gmail.com> <20120111175952.GI466@moon> <CAHGf_=oMk2YDnmsomgCc45Ei5yaSdNzRwapm7Fqm9mN7LY0FNw@mail.gmail.com> <4F0DD365.6070200@parallels.com> <20120111183115.GA28196@moon> <CAHGf_=oXsrbsXPqBw_CvZQjwdXhzhqkoWX3mRSEgMZHy44ZUKg@mail.gmail.com>
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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.
> .
>