locking/mutex: Fix HANDOFF condition

Yanfei reported that setting HANDOFF should not depend on recomputing
@first, only on @first state. Which would then give:

  if (ww_ctx || !first)
    first = __mutex_waiter_is_first(lock, &waiter);
  if (first)
    __mutex_set_flag(lock, MUTEX_FLAG_HANDOFF);

But because 'ww_ctx || !first' is basically 'always' and the test for
first is relatively cheap, omit that first branch entirely.
Reported-by: Yanfei Xu <yanfei.xu@windriver.com>
Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Reviewed-by: Waiman Long <longman@redhat.com>
Reviewed-by: Yanfei Xu <yanfei.xu@windriver.com>
Link: https://lore.kernel.org/r/20210630154114.896786297@infradead.org

kernel/locking/mutex.c

@@ -909,7 +909,6 @@ __mutex_lock_common(struct mutex *lock, long state, unsigned int subclass,
 		    struct ww_acquire_ctx *ww_ctx, const bool use_ww_ctx)
 {
 	struct mutex_waiter waiter;
-	bool first = false;
 	struct ww_mutex *ww;
 	int ret;
 
@@ -988,6 +987,8 @@ __mutex_lock_common(struct mutex *lock, long state, unsigned int subclass,
 
 	set_current_state(state);
 	for (;;) {
+		bool first;
+
 		/*
 		 * Once we hold wait_lock, we're serialized against
 		 * mutex_unlock() handing the lock off to us, do a trylock
@@ -1016,15 +1017,9 @@ __mutex_lock_common(struct mutex *lock, long state, unsigned int subclass,
 		spin_unlock(&lock->wait_lock);
 		schedule_preempt_disabled();
 
-		/*
-		 * ww_mutex needs to always recheck its position since its waiter
-		 * list is not FIFO ordered.
-		 */
-		if (ww_ctx || !first) {
-			first = __mutex_waiter_is_first(lock, &waiter);
-			if (first)
-				__mutex_set_flag(lock, MUTEX_FLAG_HANDOFF);
-		}
+		first = __mutex_waiter_is_first(lock, &waiter);
+		if (first)
+			__mutex_set_flag(lock, MUTEX_FLAG_HANDOFF);
 
 		set_current_state(state);
 		/*