[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>
This commit is contained in:
Jan Beulich 2006-01-06 00:11:48 -08:00 committed by Linus Torvalds
parent 7c4cb60e5b
commit e43d674f44

View File

@ -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);