From mboxrd@z Thu Jan 1 00:00:00 1970 From: David Miller Subject: Re: [PATCH] tcp: splice as many packets as possible at once Date: Tue, 13 Jan 2009 15:08:51 -0800 (PST) Message-ID: <20090113.150851.161423735.davem@davemloft.net> References: <49667534.5060501@zeus.com> <20090108.135515.85489589.davem@davemloft.net> <20090108222039.GC24243@1wt.eu> Mime-Version: 1.0 Content-Type: Text/Plain; charset=us-ascii Content-Transfer-Encoding: 7bit Cc: ben@zeus.com, jarkao2@gmail.com, mingo@elte.hu, linux-kernel@vger.kernel.org, netdev@vger.kernel.org, jens.axboe@oracle.com To: w@1wt.eu Return-path: Received: from 74-93-104-97-Washington.hfc.comcastbusiness.net ([74.93.104.97]:42382 "EHLO sunset.davemloft.net" rhost-flags-OK-FAIL-OK-OK) by vger.kernel.org with ESMTP id S1751661AbZAMXIu (ORCPT ); Tue, 13 Jan 2009 18:08:50 -0500 In-Reply-To: <20090108222039.GC24243@1wt.eu> Sender: netdev-owner@vger.kernel.org List-ID: From: Willy Tarreau 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.