Messages in this thread | | | Subject | Re: [RFC] per thread page reservation patch | From | Nikita Danilov <> | Date | Sat, 08 Jan 2005 00:00:00 +0300 |
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Christoph Hellwig <hch@infradead.org> writes:
> On Fri, Jan 07, 2005 at 11:48:58PM +0300, Nikita Danilov wrote: >> sufficient to create and use per-thread reservations. Using >> current->private_pages_count directly >> >> - makes API less uniform, not contained within single namespace >> (perthread_pages_*), and worse, >> >> - exhibits internal implementation detail to the user. > > Completely disagreed, hiding all the details doesn't help for such > trivial access, it's just obsfucating things. > > But looking at it the API doesn't make sense at all. Number of pages > in the thread-pool is an internal implementation detail and no caller > must look at it - think about two callers, e.g. filesystem and iscsi > initiator using it in the same thread. > > Here's an updated patch with my suggestions implemented and other goodies > such a kerneldoc comments (but completely untested so far): > > > --- 1.20/include/linux/gfp.h 2005-01-05 18:30:39 +01:00 > +++ edited/include/linux/gfp.h 2005-01-07 20:30:20 +01:00 > @@ -130,6 +130,9 @@ > #define __free_page(page) __free_pages((page), 0) > #define free_page(addr) free_pages((addr),0) > > +extern int perthread_pages_reserve(unsigned int nrpages, unsigned int gfp_mask); > +extern void perthread_pages_release(unsigned int nrpages);
I don't see how this can work.
/* make conservative estimation... */ perthread_pages_reserve(100, GFP_KERNEL);
/* actually, use only 10 pages during atomic operation */
/* want to release remaining 90 pages... */ perthread_pages_release(???);
Can you elaborate how to calculate number of pages that has to be released?
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