[PATCH v2] kcov: properly check if we are in an interrupt

[Andrey Konovalov]

in_interrupt() returns a nonzero value when we are either in an
interrupt or have bh disabled via local_bh_disable(). Since we are
interested in only ignoring coverage from actual interrupts, do a
proper check instead of just calling in_interrupt().

Signed-off-by: Andrey Konovalov <andreyknvl@xxxxxxxxxx>
---
Changes in v2:
 - Add a comment explaining why the check is open-coded.

 kernel/kcov.c | 9 ++++++++-
 1 file changed, 8 insertions(+), 1 deletion(-)

diff --git a/kernel/kcov.c b/kernel/kcov.c
index 8d44b3f..30e6d05 100644
--- a/kernel/kcov.c
+++ b/kernel/kcov.c
@@ -53,8 +53,15 @@ void notrace __sanitizer_cov_trace_pc(void)
 	/*
 	 * We are interested in code coverage as a function of a syscall inputs,
 	 * so we ignore code executed in interrupts.
+	 * The checks for whether we are in an interrupt are open-coded, because
+	 * 1. We can't use in_interrupt() here, since it also returns true
+	 *    when we are inside local_bh_disable() section.
+	 * 2. We don't want to use (in_irq() | in_serving_softirq() | in_nmi()),
+	 *    since that leads to slower generated code (three separate tests,
+	 *    one for each of the flags).
 	 */
-	if (!t || in_interrupt())
+	if (!t || (preempt_count() & (HARDIRQ_MASK | SOFTIRQ_OFFSET
+							| NMI_MASK)))
 		return;
 	mode = READ_ONCE(t->kcov_mode);
 	if (mode == KCOV_MODE_TRACE) {
-- 
2.8.0.rc3.226.g39d4020