rcu: Return early if callback is not specified

BugLink: https://bugs.launchpad.net/bugs/2122072

[ Upstream commit 33b6a1f155d627f5bd80c7485c598ce45428f74f ]

Currently the call_rcu() API does not check whether a callback
pointer is NULL. If NULL is passed, rcu_core() will try to invoke
it, resulting in NULL pointer dereference and a kernel crash.

To prevent this and improve debuggability, this patch adds a check
for NULL and emits a kernel stack trace to help identify a faulty
caller.

Signed-off-by: Uladzislau Rezki (Sony) <urezki@gmail.com>
Reviewed-by: Joel Fernandes <joelagnelf@nvidia.com>
Signed-off-by: Joel Fernandes <joelagnelf@nvidia.com>
Signed-off-by: Sasha Levin <sashal@kernel.org>
Signed-off-by: Noah Wager <noah.wager@canonical.com>
Signed-off-by: Stefan Bader <stefan.bader@canonical.com>

Author: urezki

kernel/rcu/tree.c

@@ -3055,6 +3055,10 @@ __call_rcu_common(struct rcu_head *head, rcu_callback_t func, bool lazy_in)
 	/* Misaligned rcu_head! */
 	WARN_ON_ONCE((unsigned long)head & (sizeof(void *) - 1));
 
+	/* Avoid NULL dereference if callback is NULL. */
+	if (WARN_ON_ONCE(!func))
+		return;
+
 	if (debug_rcu_head_queue(head)) {
 		/*
 		 * Probable double call_rcu(), so leak the callback.