Messages in this thread | | | From | KOSAKI Motohiro <> | Date | Wed, 11 Jan 2012 14:29:48 -0500 | Subject | Re: [RFC] on general object IDs again |
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>> > 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.
> But when we get a number of pids to dump we need the > resource affinity picture over them all. -- To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/
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