sched: Fix __schedule_bug() output when called from an interrupt

If schedule is called from an interrupt handler __schedule_bug()
will call show_regs() with the registers saved during the
interrupt handling done in do_IRQ(). This means we'll see the
registers and the backtrace for the process that was interrupted
and not the full backtrace explaining who called schedule().

This is due to 838225b ("sched: use show_regs() to improve
__schedule_bug() output", 2007-10-24) which improperly assumed
that get_irq_regs() would return the registers for the current
stack because it is being called from within an interrupt
handler. Simply remove the show_reg() code so that we dump a
backtrace for the interrupt handler that called schedule().

[ I ran across this when I was presented with a scheduling while
  atomic log with a stacktrace pointing at spin_unlock_irqrestore().
  It made no sense and I had to guess what interrupt handler could
  be called and poke around for someone calling schedule() in an
  interrupt handler. A simple test of putting an msleep() in
  an interrupt handler works better with this patch because you
  can actually see the msleep() call in the backtrace. ]

Also-reported-by: Chris Metcalf <[email protected]>
Signed-off-by: Stephen Boyd <[email protected]>
Cc: Satyam Sharma <[email protected]>
Cc: Peter Zijlstra <[email protected]>
Link: http://lkml.kernel.org/r/[email protected]
Signed-off-by: Ingo Molnar <[email protected]>

kernel/sched/core.c

@@ -3099,8 +3099,6 @@ EXPORT_SYMBOL(sub_preempt_count);
  */
 static noinline void __schedule_bug(struct task_struct *prev)
 {
-	struct pt_regs *regs = get_irq_regs();
-
 	if (oops_in_progress)
 		return;
 
@@ -3111,11 +3109,7 @@ static noinline void __schedule_bug(struct task_struct *prev)
 	print_modules();
 	if (irqs_disabled())
 		print_irqtrace_events(prev);
-
-	if (regs)
-		show_regs(regs);
-	else
-		dump_stack();
+	dump_stack();
 }
 
 /*