Re: [PATCH 00/10] foundations for reserve-based allocation

Daniel Phillips
2007-08-06T18:40:33Z

On Monday 06 August 2007 11:17, Peter Zijlstra wrote:
> lim_{n -> inf} (2^(n+1)/((2^n)+1)) = 2^lim_{n -> inf} ((n+1)-n) = 2^1
= 2

Glad I asked :-)

> > Patch [3/10] adds a new field to struct page.
>
> No it doesn't.

True. It is not immediately obvious from the declaration that the
overloaded field is always freed up before anybody else needs to use
the union.

> > I do not think this is
> > necessary. Allocating a page from reserve does not make it
> > special. All we care about is that the total number of pages taken
> > out of reserve is balanced by the total pages freed by a user of
> > the reserve.
>
> And how do we know a page was taken out of the reserves?
>
> This is done by looking at page->reserve (overload of page->index)
> and this value can be destroyed as soon as its observed. It is in a
> sense an extra return value.

Ah I see. I used to let alloc_pages fail then repeat the allocation
with __GFP_MEMALLOC set, which was easy but stupidly repetitious. Your
technique is better, though returning the status in the page still
looks a little funny. This is really about saving a page flag, no?

> > We do care about slab fragmentation in the sense that a slab page
> > may be pinned in the slab by an unprivileged allocation and so that
> > page may never be returned to the global page reserve.
>
> A slab page obtained from the reseserve will never serve an object to
> an unprivilidged allocation.
>
> > One way to solve this is
> > to have a per slabpage flag indicating the page came from reserve,
> > and prevent mixing of privileged and unprivileged allocations on
> > such a page.
>
> is done.

Serves me right for not reading that bit. So the score for that round
is: peterz 3, phillips 0 ;-)

Regards,

Daniel