From mboxrd@z Thu Jan  1 00:00:00 1970
From: valdis.kletnieks@vt.edu (valdis.kletnieks at vt.edu)
Date: Mon, 04 Jun 2018 12:10:07 -0400
Subject: develoment workflow: how to avoid duplicate work ?
In-Reply-To: <20180604033303.GX14048@gavran.carpriv.carnet.hr>
References: <20180529030706.GD1845@hle-laptop.local>
 <80132.1527564735@turing-police.cc.vt.edu>
 <20180530025657.GB1919@hle-laptop.local>
 <22501.1527692619@turing-police.cc.vt.edu>
 <20180531014432.GA3722@hle-laptop.local>
 <20180603192336.GV14048@gavran.carpriv.carnet.hr>
 <20180603222556.GB20060@hle-laptop.local>
 <20180604033303.GX14048@gavran.carpriv.carnet.hr>
Message-ID: <12368.1528128607@turing-police.cc.vt.edu>
To: kernelnewbies@lists.kernelnewbies.org
List-Id: kernelnewbies.lists.kernelnewbies.org

On Mon, 04 Jun 2018 05:33:03 +0200, Valentin Vidic said:
> On Sun, Jun 03, 2018 at 06:25:56PM -0400, Hugo Lefeuvre wrote:
> > The vfs documentation states: release() is "called when the last
> > reference to an open file is closed".
> > 
> > Let's say we have a program with threads T1 and T2.
> > 
> > - T1 calls ioctl on a file descriptor FD.
> > - (on another processor) T2 closes FD.
> > 
> > Since the last reference to FD was closed by T2, release is called.

That's subtly wrong.  T2 releases its reference to the file descriptor.

> > But while release is being called, the ioctl call from T1 may still
> > be running, right ?

Remember that ioctl needs an open FD as well - so the ioctl() grabs its own
reference, and then *that* reference to the file descriptor stays in place at
least until the ioctl() return.  At *that* point, the reference count goes to
zero and the file is actually closed.