kernel/locks: Tweak mutex_unlock KDEBUG semantics slightly.

Previously, both the "old" owner and the "new" lock owner set
the lock holder to the "new" lock holder. Now the old owner does
it before calling unblock(), and we check in the lock() functions
that we are indeed the new owner of the lock.

This may affect what the panic is for #15211 and its duplicates:
I have a suspicion that these threads are getting unblocked
when they have no business being unblocked, and this should
catch that condition.

src/system/kernel/locks/lock.cpp

@@ -781,8 +781,9 @@ _mutex_lock(mutex* lock, void* _locker)
 
 	status_t error = thread_block();
 #if KDEBUG
-	if (error == B_OK)
-		atomic_set(&lock->holder, waiter.thread->id);
+	if (error == B_OK) {
+		ASSERT(lock->holder == waiter.thread->id);
+	}
 #endif
 	return error;
 }
@@ -813,23 +814,19 @@ _mutex_unlock(mutex* lock)
 		lock->waiters = waiter->next;
 		if (lock->waiters != NULL)
 			lock->waiters->last = waiter->last;
-#if KDEBUG
-		thread_id unblockedThread = waiter->thread->id;
-#endif
-
-		// unblock thread
-		thread_unblock(waiter->thread, B_OK);
 
 #if KDEBUG
 		// Already set the holder to the unblocked thread. Besides that this
 		// actually reflects the current situation, setting it to -1 would
 		// cause a race condition, since another locker could think the lock
 		// is not held by anyone.
-		lock->holder = unblockedThread;
+		lock->holder = waiter->thread->id;
 #endif
+
+		// unblock thread
+		thread_unblock(waiter->thread, B_OK);
 	} else {
-		// We've acquired the spinlock before the locker that is going to wait.
-		// Just mark the lock as released.
+		// Nobody is waiting to acquire this lock. Just mark it as released.
 #if KDEBUG
 		lock->holder = -1;
 #else
@@ -907,7 +904,7 @@ _mutex_lock_with_timeout(mutex* lock, uint32 timeoutFlags, bigtime_t timeout)
 
 	if (error == B_OK) {
 #if KDEBUG
-		lock->holder = waiter.thread->id;
+		ASSERT(lock->holder == waiter.thread->id);
 #endif
 	} else {
 		locker.Lock();