Re: [RFC] per thread page reservation patch

[Nikita Danilov <nikita@xxxxxxxxxxxxx>]

Christoph Hellwig <hch@xxxxxxxxxxxxx> 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?

Nikita.
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