[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>

diff --git a/arch/i386/kernel/traps.c b/arch/i386/kernel/traps.c
index ab0e9430f775cfd22cce9b39844b6e8f1b147798..bb36a989db7789976495c69a7c4703b077142873 100644 (file)
--- 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);