Commit b274c0bb authored by Andrey Konovalov's avatar Andrey Konovalov Committed by Linus Torvalds

kcov: properly check if we are in an interrupt

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().

As a result of this change, kcov will start to collect coverage from
within local_bh_disable()/local_bh_enable() sections.

Link: default avatarAndrey Konovalov <>
Acked-by: default avatarDmitry Vyukov <>
Cc: Nicolai Stange <>
Cc: Andrey Ryabinin <>
Cc: Kees Cook <>
Cc: James Morse <>
Cc: Vegard Nossum <>
Cc: Quentin Casasnovas <>
Signed-off-by: default avatarAndrew Morton <>
Signed-off-by: default avatarLinus Torvalds <>
parent 86d9f485
......@@ -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
mode = READ_ONCE(t->kcov_mode);
if (mode == KCOV_MODE_TRACE) {
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