panic: fix a possible deadlock in panic()

panic_lock is meant to ensure that panic processing takes place only on
one cpu; if any of the other cpus encounter a panic, they will spin
waiting to be shut down.

However, this causes a regression in this scenario:

1. Cpu 0 encounters a panic and acquires the panic_lock
   and proceeds with the panic processing.
2. There is an interrupt on cpu 0 that also encounters
   an error condition and invokes panic.
3. This second invocation fails to acquire the panic_lock
   and enters the infinite while loop in panic_smp_self_stop.

Thus all panic processing is stopped, and the cpu is stuck for eternity
in the while(1) inside panic_smp_self_stop.

To address this, disable local interrupts with local_irq_disable before
acquiring the panic_lock.  This will prevent interrupt handlers from
executing during the panic processing, thus avoiding this particular
problem.
Signed-off-by: NVikram Mulukutla <markivx@codeaurora.org>
Reviewed-by: NStephen Boyd <sboyd@codeaurora.org>
Cc: Michael Holzheu <holzheu@linux.vnet.ibm.com>
Signed-off-by: NAndrew Morton <akpm@linux-foundation.org>
Signed-off-by: NLinus Torvalds <torvalds@linux-foundation.org>

kernel/panic.c

@@ -74,6 +74,14 @@ void panic(const char *fmt, ...)
 	long i, i_next = 0;
 	int state = 0;
 
+	/*
+	 * Disable local interrupts. This will prevent panic_smp_self_stop
+	 * from deadlocking the first cpu that invokes the panic, since
+	 * there is nothing to prevent an interrupt handler (that runs
+	 * after the panic_lock is acquired) from invoking panic again.
+	 */
+	local_irq_disable();
+
 	/*
 	 * It's possible to come here directly from a panic-assertion and
 	 * not have preempt disabled. Some functions called from here want