Messages in this thread | | | From | Eric Dumazet <> | Subject | Re: [PATCH 2/5] fdtable: Make fdarray and fdsets equal in size. | Date | Tue, 10 Oct 2006 19:12:56 +0200 |
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On Friday 06 October 2006 06:51, Vadim Lobanov wrote:
> - nfds = max_t(int, 8 * L1_CACHE_BYTES, roundup_pow_of_two(nr + 1)); > - if (nfds > NR_OPEN) > - nfds = NR_OPEN; > - > - new_openset = alloc_fdset(nfds); > - new_execset = alloc_fdset(nfds); > - if (!new_openset || !new_execset) > - goto out; > - fdt->open_fds = new_openset; > - fdt->close_on_exec = new_execset; > - fdt->max_fdset = nfds; > - > nfds = NR_OPEN_DEFAULT; > /* > * Expand to the max in easy steps, and keep expanding it until > @@ -271,15 +254,21 @@ static struct fdtable *alloc_fdtable(int > nfds = NR_OPEN; > } > } while (nfds <= nr);
If I understand well, we may allocate very small fdset, while previous minimum size was L1_CACHE_BYTES bytes. (512 bits for a 64 bytes cache line)
If you check commit 0c9e63fd38a2fb2181668a0cdd622a3c23cfd567, (http://www.kernel.org/git/?p=linux/kernel/git/torvalds/linux-2.6.git;a=commit;h=0c9e63fd38a2fb2181668a0cdd622a3c23cfd567 ) you'll find this comment of mine :
3) Reduce size of allocated fdset. Currently two full pages are allocated, that is 32768 bits on x86 for example, and way too much. The minimum is now L1_CACHE_BYTES.
This minimum is mandatory to be sure two tasks wont share the same cache line to store their fdset (and possibly do lot of cache line ping pongs)
Eric
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