[PATCH] i386: don't blindly enable interrupts in die()

Rather than blindly re-enabling interrupts in die(), save their state
upon entry and then restore that state.

If the kernel is in really bad condition and faults with interrupts disabled,
re-enabling them in die() may cause even more trouble, implying more chances
of data corruption.

Signed-off-by: Andrew Morton <akpm@osdl.org>
Signed-off-by: Linus Torvalds <torvalds@osdl.org>

1 files changed, 5 insertions, 2 deletions

diff --git a/arch/i386/kernel/traps.c b/arch/i386/kernel/traps.c
index ab0e9430f775..bb36a989db77 100644
--- a/arch/i386/kernel/traps.c
+++ b/arch/i386/kernel/traps.c
@@ -306,14 +306,17 @@ void die(const char * str, struct pt_regs * regs, long err)
                 .lock_owner_depth =     0
         };
         static int die_counter;
+        unsigned long flags;
 
         if (die.lock_owner != raw_smp_processor_id()) {
                 console_verbose();
-                spin_lock_irq(&die.lock);
+                spin_lock_irqsave(&die.lock, flags);
                 die.lock_owner = smp_processor_id();
                 die.lock_owner_depth = 0;
                 bust_spinlocks(1);
         }
+        else
+                local_save_flags(flags);
 
         if (++die.lock_owner_depth < 3) {
                 int nl = 0;
@@ -340,7 +343,7 @@ void die(const char * str, struct pt_regs * regs, long err)
 
         bust_spinlocks(0);
         die.lock_owner = -1;
-        spin_unlock_irq(&die.lock);
+        spin_unlock_irqrestore(&die.lock, flags);
 
         if (kexec_should_crash(current))
                 crash_kexec(regs);