Messages in this thread | | | From | Vadim Lobanov <> | Subject | Re: [PATCH 2/5] fdtable: Make fdarray and fdsets equal in size. | Date | Tue, 10 Oct 2006 11:36:34 -0700 |
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On Tuesday 10 October 2006 10:12, Eric Dumazet wrote: > 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=com >mit;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)
Good point. This should be easy to fix -- we can simply bump up the allocation 'nr' when calling alloc_fdmem() for the fdset. Since we won't cross a page size, the rest of the code will still stay the same.
I'll code this up once I track down why other folks are experiencing oopses.
> Eric
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