netdev - Re: [PATCH] tcp: splice as many packets as possible at once

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Message-Id: <20090113.150851.161423735.davem@davemloft.net>
Date:	Tue, 13 Jan 2009 15:08:51 -0800 (PST)
From:	David Miller <davem@...emloft.net>
To:	w@....eu
Cc:	ben@...s.com, jarkao2@...il.com, mingo@...e.hu,
	linux-kernel@...r.kernel.org, netdev@...r.kernel.org,
	jens.axboe@...cle.com
Subject: Re: [PATCH] tcp: splice as many packets as possible at once

From: Willy Tarreau <w@....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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