From mboxrd@z Thu Jan 1 00:00:00 1970 From: Ian Campbell Subject: Re: "don't bugger nd->seq" seems to break umount sometimes Date: Tue, 19 May 2015 10:47:27 +0100 Message-ID: <1432028847.12989.62.camel@citrix.com> References: <21824.50281.405589.957467@mariner.uk.xensource.com> <20150429122139.GN889@ZenIV.linux.org.uk> <55418219.8080700@huawei.com> <1432024439.12989.7.camel@citrix.com> <555AFCEA.5020102@huawei.com> Mime-Version: 1.0 Content-Type: text/plain; charset="us-ascii" Content-Transfer-Encoding: 7bit Return-path: In-Reply-To: <555AFCEA.5020102@huawei.com> List-Unsubscribe: , List-Post: List-Help: List-Subscribe: , Sender: xen-devel-bounces@lists.xen.org Errors-To: xen-devel-bounces@lists.xen.org To: Zefan Li Cc: xen-tools@packages.debian.org, xen-devel@lists.xensource.com, stefano.stabellini@eu.citrix.com, Andrew Cooper , Ian Jackson , Al Viro , Boris Ostrovsky , David Vrabel List-Id: xen-devel@lists.xenproject.org On Tue, 2015-05-19 at 17:05 +0800, Zefan Li wrote: > On 2015/5/19 16:33, Ian Campbell wrote: > > On Thu, 2015-04-30 at 09:15 +0800, Zefan Li wrote: > >> On 2015/4/29 20:21, Al Viro wrote: > >>> On Wed, Apr 29, 2015 at 12:45:45PM +0100, Ian Jackson wrote: > >>> > >>>> The symptoms are that `umount' fails with EBUSY, > >>> > >>> [lizf: Backported to 3.4: > >>> - remove the changes to follow_link() as it doesn't call set_root()] > >>> > >>> looks dubious - I don't have -stable in front of me, but set_root() in > >>> follow_link() had migrated from __vfs_follow_link(), so could you try > >>> (tr '#' '\t' | ed fs/namei.c) <<'EOF' > >>> /__vfs_follow_link/ > >>> /set_root/s/^/#/ > >>> i > >>> ##if (!nd->root.mnt) > >>> . > >>> wq > >>> EOF > >>> and see if it helps? > >>> . > >>> > >> > >> My fault. I just checked 3.2.y, and it made the right change. > > > > Is there going to be a 3.4.108 with a fix for this? > > > > definitely. Great. Do you have a timeline in mind? We have a daily automated test which has been hitting this issue for a while, I can continue to ignore the associated cron spam of course ;-) Ian.