Re: [PATCH 2/5] fdtable: Make fdarray and fdsets equal in size.

[Vadim Lobanov <vlobanov@speakeasy.net>]

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

-- Vadim Lobanov