On Tue, Sep 08, 2020 at 05:33:40PM +0200, Martin Wilck wrote:
> On Tue, 2020-09-08 at 10:14 -0400, Michael S. Tsirkin wrote:
> > On Mon, Aug 31, 2020 at 02:37:26PM +0200, Laurent Vivier wrote:
> > > On 28/08/2020 23:34, Martin Wilck wrote:
> > > > On Wed, 2020-08-26 at 08:26 -0400, Michael S. Tsirkin wrote:
> > > > > On Tue, Aug 11, 2020 at 04:42:32PM +0200, Laurent Vivier wrote:
> > > > > > On 11/08/2020 16:28, mwilck@xxxxxxxx wrote:
> > > > > > > From: Martin Wilck <mwilck@xxxxxxxx>
> > > > > > >
> > > > > > > If a program opens /dev/hwrng with O_NONBLOCK and uses
> > > > > > > poll() and
> > > > > > > non-blocking read() to retrieve random data, it ends up in
> > > > > > > a
> > > > > > > tight
> > > > > > > loop with poll() always returning POLLIN and read()
> > > > > > > returning
> > > > > > > EAGAIN.
> > > > > > > This repeats forever until some process makes a blocking
> > > > > > > read()
> > > > > > > call.
> > > > > > > The reason is that virtio_read() always returns 0 in non-
> > > > > > > blocking
> > > > > > > mode,
> > > > > > > even if data is available. Worse, it fetches random data
> > > > > > > from the
> > > > > > > hypervisor after every non-blocking call, without ever
> > > > > > > using this
> > > > > > > data.
> > > > > > >
> > > > > > > The following test program illustrates the behavior and can
> > > > > > > be
> > > > > > > used
> > > > > > > for testing and experiments. The problem will only be seen
> > > > > > > if all
> > > > > > > tasks use non-blocking access; otherwise the blocking reads
> > > > > > > will
> > > > > > > "recharge" the random pool and cause other, non-blocking
> > > > > > > reads to
> > > > > > > succeed at least sometimes.
> > > > > > >
> > > > > > > /* Whether to use non-blocking mode in a task, problem
> > > > > > > occurs if
> > > > > > > CONDITION is 1 */
> > > > > > > //#define CONDITION (getpid() % 2 != 0)
> > > > > > >
> > > > > > > static volatile sig_atomic_t stop;
> > > > > > > static void handler(int sig __attribute__((unused))) { stop
> > > > > > > = 1;
> > > > > > > }
> > > > > > >
> > > > > > > static void loop(int fd, int sec)
> > > > > > > {
> > > > > > > struct pollfd pfd = { .fd = fd, .events = POLLIN, };
> > > > > > > unsigned long errors = 0, eagains = 0, bytes = 0, succ
> > > > > > > = 0;
> > > > > > > int size, rc, rd;
> > > > > > >
> > > > > > > srandom(getpid());
> > > > > > > if (CONDITION && fcntl(fd, F_SETFL, fcntl(fd, F_GETFL)
> > > > > > > |
> > > > > > > O_NONBLOCK) == -1)
> > > > > > > perror("fcntl");
> > > > > > > size = MINBUFSIZ + random() % (MAXBUFSIZ - MINBUFSIZ +
> > > > > > > 1);
> > > > > > >
> > > > > > > for(;;) {
> > > > > > > char buf[size];
> > > > > > >
> > > > > > > if (stop)
> > > > > > > break;
> > > > > > > rc = poll(&pfd, 1, sec);
> > > > > > > if (rc > 0) {
> > > > > > > rd = read(fd, buf, sizeof(buf));
> > > > > > > if (rd == -1 && errno == EAGAIN)
> > > > > > > eagains++;
> > > > > > > else if (rd == -1)
> > > > > > > errors++;
> > > > > > > else {
> > > > > > > succ++;
> > > > > > > bytes += rd;
> > > > > > > write(1, buf, sizeof(buf));
> > > > > > > }
> > > > > > > } else if (rc == -1) {
> > > > > > > if (errno != EINTR)
> > > > > > > perror("poll");
> > > > > > > break;
> > > > > > > } else
> > > > > > > fprintf(stderr, "poll: timeout\n");
> > > > > > > }
> > > > > > > fprintf(stderr,
> > > > > > > "pid %d %sblocking, bufsize %d, %d seconds, %lu
> > > > > > > bytes
> > > > > > > read, %lu success, %lu eagain, %lu errors\n",
> > > > > > > getpid(), CONDITION ? "non-" : "", size, sec,
> > > > > > > bytes,
> > > > > > > succ, eagains, errors);
> > > > > > > }
> > > > > > >
> > > > > > > int main(void)
> > > > > > > {
> > > > > > > int fd;
> > > > > > >
> > > > > > > fork(); fork();
> > > > > > > fd = open("/dev/hwrng", O_RDONLY);
> > > > > > > if (fd == -1) {
> > > > > > > perror("open");
> > > > > > > return 1;
> > > > > > > };
> > > > > > > signal(SIGALRM, handler);
> > > > > > > alarm(SECONDS);
> > > > > > > loop(fd, SECONDS);
> > > > > > > close(fd);
> > > > > > > wait(NULL);
> > > > > > > return 0;
> > > > > > > }
> > > > > > >
> > > > > > > void loop(int fd)
> > > > > > > {
> > > > > > > struct pollfd pfd0 = { .fd = fd, .events = POLLIN,
> > > > > > > };
> > > > > > > int rc;
> > > > > > > unsigned int n;
> > > > > > >
> > > > > > > for (n = LOOPS; n > 0; n--) {
> > > > > > > struct pollfd pfd = pfd0;
> > > > > > > char buf[SIZE];
> > > > > > >
> > > > > > > rc = poll(&pfd, 1, 1);
> > > > > > > if (rc > 0) {
> > > > > > > int rd = read(fd, buf,
> > > > > > > sizeof(buf));
> > > > > > >
> > > > > > > if (rd == -1)
> > > > > > > perror("read");
> > > > > > > else
> > > > > > > printf("read %d bytes\n",
> > > > > > > rd);
> > > > > > > } else if (rc == -1)
> > > > > > > perror("poll");
> > > > > > > else
> > > > > > > fprintf(stderr, "timeout\n");
> > > > > > >
> > > > > > > }
> > > > > > > }
> > > > > > >
> > > > > > > int main(void)
> > > > > > > {
> > > > > > > int fd;
> > > > > > >
> > > > > > > fd = open("/dev/hwrng", O_RDONLY|O_NONBLOCK);
> > > > > > > if (fd == -1) {
> > > > > > > perror("open");
> > > > > > > return 1;
> > > > > > > };
> > > > > > > loop(fd);
> > > > > > > close(fd);
> > > > > > > return 0;
> > > > > > > }
> > > > > > >
> > > > > > > This can be observed in the real word e.g. with nested
> > > > > > > qemu/KVM
> > > > > > > virtual
> > > > > > > machines, if both the "outer" and "inner" VMs have a
> > > > > > > virtio-rng
> > > > > > > device.
> > > > > > > If the "inner" VM requests random data, qemu running in the
> > > > > > > "outer" VM
> > > > > > > uses this device in a non-blocking manner like the test
> > > > > > > program
> > > > > > > above.
> > > > > > >
> > > > > > > Fix it by returning available data if a previous hypervisor
> > > > > > > call
> > > > > > > has
> > > > > > > completed. I tested this patch with the program above, and
> > > > > > > with
> > > > > > > rng-tools.
> > > > > > >
> > > > > > > v2 -> v3: Simplified the implementation as suggested by
> > > > > > > Laurent
> > > > > > > Vivier
> > > > > > >
> > > > > > > Signed-off-by: Martin Wilck <mwilck@xxxxxxxx>
> > > > > > > ---
> > > > > > > drivers/char/hw_random/virtio-rng.c | 4 ++--
> > > > > > > 1 file changed, 2 insertions(+), 2 deletions(-)
> > > > > > >
> > > > > > > diff --git a/drivers/char/hw_random/virtio-rng.c
> > > > > > > b/drivers/char/hw_random/virtio-rng.c
> > > > > > > index a90001e02bf7..8eaeceecb41e 100644
> > > > > > > --- a/drivers/char/hw_random/virtio-rng.c
> > > > > > > +++ b/drivers/char/hw_random/virtio-rng.c
> > > > > > > @@ -65,7 +65,7 @@ static int virtio_read(struct hwrng *rng,
> > > > > > > void
> > > > > > > *buf, size_t size, bool wait)
> > > > > > > register_buffer(vi, buf, size);
> > > > > > > }
> > > > > > >
> > > > > > > - if (!wait)
> > > > > > > + if (!wait && !completion_done(&vi->have_data))
> > > > > > > return 0;
> > > > > > >
> > > > > > > ret = wait_for_completion_killable(&vi->have_data);
> > > > > > > @@ -74,7 +74,7 @@ static int virtio_read(struct hwrng *rng,
> > > > > > > void
> > > > > > > *buf, size_t size, bool wait)
> > > > > > >
> > > > > > > vi->busy = false;
> > > > > > >
> > > > > > > - return vi->data_avail;
> > > > > > > + return min_t(size_t, size, vi->data_avail);
> > > > > > > }
> > > > > > >
> > > > > > > static void virtio_cleanup(struct hwrng *rng)
> > > > > > >
> > > > > >
> > > > > > Reviewed-by: Laurent Vivier <lvivier@xxxxxxxxxx>
> > > > >
> > > > > Laurent didn't we agree the real fix is private buffers in the
> > > > > driver,
> > > > > and copying out from there?
> > > > >
> > > >
> > > > Can we perhaps proceed with this for now? AFAICS the private
> > > > buffer
> > > > implementation would be a larger effort, while we have the issues
> > > > with
> > > > nested VMs getting no entropy today.
> > > >
> > >
> > > I agree. I think it's important to have a simple and quick fix for
> > > the
> > > problem reported by Martin.
> > >
> > > We need the private buffers but not sure how long it will take to
> > > have
> > > them included in the kernel and how many new bugs will be
> > > introduced
> > > doing that as the code is hard to understand and the core is shared
> > > with
> > > several other hardware backends that can be impacted by the changes
> > > needed.
> > >
> > > Thanks,
> > > Laurent
> >
> > However I am not sure with the patch applies we never return
> > the same buffer to userspace twice, e.g. if one is
> > non blocking another blocking. Doing that would be a bug.
> >
>
> As Laurent mentioned in
> https://lists.gnu.org/archive/html/qemu-devel/2020-08/msg02039.html,
> there are only 2 different buffers that may be passed to virtio_read(),
> rng_buffer and rng_fillbuf.
> The latter is only used in blocking mode.
>
> AFAICS there's just one problematic situation:
>
> 1 a user space process reads random data without blocking and runs
> register_buffer(), gets no data, releases reading_mutex
> 2 the hwrng kthread grabs the mutex and makes a sync call, vi->busy is
> still set, so no new completion is initialized.
> 3 hwrng calls wait_for_completion_killable() and sees the completion
> that had been initialized by the user space process previously,
> 4 hwrng "thinks" it got some positive randomness, but random data have
> actually been written into rng_buffer, not rng_fillbuff.
>
> This is indeed bad, but it can happen with the current code as well.
> Actually, it's more likely to happen with the current code, because
> asynchronous callers might hang forever trying to get entropy,
> making this scenario more likely (if there's a process, like nested
> qemu, that would keep calling . So this wouldn't be a regression caused
> by my patch, AFAICT.
>
> How can we avoid this problem entirely? A) With private buffers, of
> course. B) Another, a bit hackish, approach would be to remember the
> active "buffer" pointer in virtio_rng, and restart the IO when a
> another buffer is passed down. C) Finally, we could modify
> virtio_read() such that blocking calls always re-initialize the buffer;
> they'd then have to wait for a potential already running IO from a
> previous, non-blocking access to finish first.
>
> But I believe this is something which could (and should) be done
> independently. Alternatively, I could add B) or C). A) I'd rather leave
> to Laurent.
>
> Regards,
> Martin
Of the simple solutions, C seems cleanest.
Laurent, any interest in working on A meanwhile?
--
MST
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