Date: Tue, 20 Nov 2012 22:28:56 +0800
From: Fengguang Wu <fengguang.wu@intel.com>
To: Jaegeuk Hanse <jaegeuk.hanse@gmail.com>
Cc: Claudio Freire <klaussfreire@gmail.com>,
        Andrew Morton <akpm@linux-foundation.org>,
        linux-kernel@vger.kernel.org,
        Linux Memory Management List <linux-mm@kvack.org>
Subject: Re: fadvise interferes with readahead
Message-ID: <20121120142856.GA19467@localhost>
References: <CAGTBQpaDR4+V5b1AwAVyuVLu5rkU=Wc1WeUdLu5ag=WOk5oJzQ@mail.gmail.com>
 <20121120080427.GA11019@localhost>
 <50AB8396.4040504@gmail.com>
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> >Yes. The kernel readahead code by design will outperform simple
> >fadvise in the case of clustered random reads. Imagine the access
> >pattern 1, 3, 2, 6, 4, 9. fadvise will trigger 6 IOs literally. While
> 
> You mean it will trigger 6 IOs in the POSIX_FADV_RANDOM case or
> POSIX_FADV_WILLNEED case?

Yes. However note that I'm assuming 1-page sized and prefetch depth
fadvise(POSIX_FADV_WILLNEED) calls in this example. Given more
prefetch depth or good timing, there will be possibility for IO
requests (eg. 3 and 2) be merged at block layer.

> >kernel readahead will likely trigger 3 IOs for 1, 3, 2-9. Because on
> >the page miss for 2, it will detect the existence of history page 1
> >and do readahead properly. For hard disks, it's mainly the number of
> 
> If the first IO read 1, it will call page_cache_sync_read() since
> cache miss,
> if (offset - (ra->prev_pos) >> PAGE_CACHE_SHIFT) <= 1UL)
>     goto initial_readahead;
> If the initial_readahead will be called? Because offset is equal to
> 1 and ra->prev_pos is equal to 0. If my assume is true, 2 also will
> be readahead.

ra->prev_pos is initialized to -1 in file_ra_state_init(), so that if
the very first read is on page 0, it will trigger readahead.

Sorry I gave a confusing example. We may as well use 1001, 1003, 1002,
1006, 1004, 1009 as the example numbers.

Thanks,
Fengguang
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