panic: fix stack dump print on direct call to panic()

Commit 6e6f0a1 ("panic: don't print redundant backtraces on oops")
causes a regression where no stack trace will be printed at all for the
case where kernel code calls panic() directly while not processing an
oops, and of course there are 100's of instances of this type of call.

The original commit executed the check (!oops_in_progress), but this will
always be false because just before the dump_stack() there is a call to
bust_spinlocks(1), which does the following:

  void __attribute__((weak)) bust_spinlocks(int yes)
  {
	if (yes) {
		++oops_in_progress;

The proper way to resolve the problem that original commit tried to
solve is to avoid printing a stack dump from panic() when the either of
the following conditions is true:

  1) TAINT_DIE has been set (this is done by oops_end())
     This indicates and oops has already been printed.
  2) oops_in_progress > 1
     This guards against the rare case where panic() is invoked
     a second time, or in between oops_begin() and oops_end()

Signed-off-by: Jason Wessel <[email protected]>
Cc: Andi Kleen <[email protected]>
Cc: <[email protected]>	[3.3+]
Signed-off-by: Andrew Morton <[email protected]>
Signed-off-by: Linus Torvalds <[email protected]>

kernel/panic.c

@@ -97,7 +97,7 @@ void panic(const char *fmt, ...)
 	/*
 	 * Avoid nested stack-dumping if a panic occurs during oops processing
 	 */
-	if (!oops_in_progress)
+	if (!test_taint(TAINT_DIE) && oops_in_progress <= 1)
 		dump_stack();
 #endif