From: Hou Tao <houtao1@xxxxxxxxxx>
bpf_send_signal_common() uses preemptible() to check whether or not the
current context is preemptible. If it is preemptible, it will use
irq_work to send the signal asynchronously instead of trying to hold a
spin-lock, because spin-lock is sleepable under PREEMPT_RT.
However, preemptible() depends on CONFIG_PREEMPT_COUNT. When
CONFIG_PREEMPT_COUNT is turned off (e.g., CONFIG_PREEMPT_VOLUNTARY=y),
!preemptible() will be evaluated as 1 and bpf_send_signal_common() will
use irq_work unconditionally.
Fix it by unfolding "!preemptible()" and using "preempt_count() != 0 ||
irqs_disabled()" instead.
Fixes: 87c544108b61 ("bpf: Send signals asynchronously if !preemptible")
Signed-off-by: Hou Tao <houtao1@xxxxxxxxxx>
---
Hi,
I had reported the problem at [1], however, it seems it got lost or was
ignored. Therefore, I sent it out myself.
[1]: https://lore.kernel.org/all/94fcfa71-51c1-2130-3452-ec58aaef94d0@xxxxxxxxxxxxxxx/
kernel/trace/bpf_trace.c | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/kernel/trace/bpf_trace.c b/kernel/trace/bpf_trace.c
index 71c1c02ca7a3e..2e6b525904d31 100644
--- a/kernel/trace/bpf_trace.c
+++ b/kernel/trace/bpf_trace.c
@@ -842,7 +842,7 @@ static int bpf_send_signal_common(u32 sig, enum pid_type type, struct task_struc
if (unlikely(is_global_init(task)))
return -EPERM;
- if (!preemptible()) {
+ if (preempt_count() != 0 || irqs_disabled()) {
/* Do an early check on signal validity. Otherwise,
* the error is lost in deferred irq_work.
*/
--
2.29.2
