Messages in this thread | | | Date | Tue, 13 Jan 2009 15:08:51 -0800 (PST) | Subject | Re: [PATCH] tcp: splice as many packets as possible at once | From | David Miller <> |
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From: Willy Tarreau <w@1wt.eu> Date: Thu, 8 Jan 2009 23:20:39 +0100
> On Thu, Jan 08, 2009 at 01:55:15PM -0800, David Miller wrote: > > I'm not applying this until someone explains to me why > > we should remove this test from the splice receive but > > keep it in the tcp_recvmsg() code where it has been > > essentially forever. > > In my opinion, the code structure is different between both functions. In > tcp_recvmsg(), we test for it if (copied > 0), where copied is the sum of > all data which have been processed since the entry in the function. If we > removed the test here, we could not break out of the loop once we have > copied something. In tcp_splice_read(), the test is still present in the > (!ret) code path, where ret is the last number of bytes processed, so the > test is still performed regardless of what has been previously transferred. > > So in summary, in tcp_splice_read without this test, we get back to the > top of the loop, and if __tcp_splice_read() returns 0, then we break out > of the loop.
Ok I see what you're saying, the !timeo check is only necessary when the receive queue has been exhausted.
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