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bpf: Use preempt_count() directly in bpf_send_signal_common()
[ Upstream commit b4a8b5bba712a711d8ca1f7d04646db63f9c88f5 ]
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@huawei.com>
Link: https://lore.kernel.org/r/20250220042259.1583319-1-houtao@huaweicloud.com
Signed-off-by: Alexei Starovoitov <ast@kernel.org>
Signed-off-by: Sasha Levin <sashal@kernel.org>
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1 changed files with 1 additions and 1 deletions
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@ -653,7 +653,7 @@ BPF_CALL_1(bpf_send_signal, u32, sig)
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if (unlikely(is_global_init(current)))
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return -EPERM;
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if (!preemptible()) {
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if (preempt_count() != 0 || irqs_disabled()) {
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/* Do an early check on signal validity. Otherwise,
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* the error is lost in deferred irq_work.
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*/
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