bpf: Use preempt_count() directly in bpf_send_signal_common()

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: 87c5441 ("bpf: Send signals asynchronously if !preemptible")
Signed-off-by: Hou Tao <houtao1@huawei.com>
Link: https://lore.kernel.org/r/20250220042259.1583319-1-houtao@huaweicloud.com
Signed-off-by: Alexei Starovoitov <ast@kernel.org>

kernel/trace/bpf_trace.c

@@ -843,7 +843,7 @@ 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.
 		 */