Consider two threads calling read() on the same uinput-fd, both
non-blocking. Assume there is data-available so both will simultaneously
pass:
udev->head == udev->tail
Then the first thread goes to sleep and the second one pops the message
from the queue. Now assume udev->head == udev->tail. If the first thread
wakes up it will call wait_event_*() and sleep in the waitq. This
effectively turns the non-blocking FD into a blocking one.
We fix this by never calling wait_event_*() for non-blocking FDs hence we
will never sleep in the waitq here.
Signed-off-by: David Herrmann <dh.herrmann@xxxxxxxxxxxxxx>
Acked-by: Aristeu Rozanski <aris@xxxxxxxxx>
---
drivers/input/misc/uinput.c | 14 ++++++++------
1 file changed, 8 insertions(+), 6 deletions(-)
diff --git a/drivers/input/misc/uinput.c b/drivers/input/misc/uinput.c
index 7360568..1526814 100644
--- a/drivers/input/misc/uinput.c
+++ b/drivers/input/misc/uinput.c
@@ -460,13 +460,15 @@ static ssize_t uinput_read(struct file *file, char __user *buffer, size_t count,
if (udev->state != UIST_CREATED)
return -ENODEV;
- if (udev->head == udev->tail && (file->f_flags & O_NONBLOCK))
- return -EAGAIN;
-
- retval = wait_event_interruptible(udev->waitq,
+ if (file->f_flags & O_NONBLOCK) {
+ if (udev->head == udev->tail)
+ return -EAGAIN;
+ } else {
+ retval = wait_event_interruptible(udev->waitq,
udev->head != udev->tail || udev->state != UIST_CREATED);
- if (retval)
- return retval;
+ if (retval)
+ return retval;
+ }
retval = mutex_lock_interruptible(&udev->mutex);
if (retval)
--
1.7.9.4
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