Ingo Molnar wrote:
+
+/*
+ * Spinlock based trylock, we take the spinlock and check whether we
+ * can get the lock:
+ */
+static inline int __mutex_trylock_nonatomic(atomic_t *lock_count)
+{
+ struct mutex *lock = container_of(lock_count, struct mutex, count);
+ int prev;
+
+ spin_lock_mutex(&lock->wait_lock);
+
+ prev = atomic_xchg(&lock->count, -1);
+ if (likely(prev == 1))
+ debug_mutex_set_owner(lock, current_thread_info() __RET_IP__);
+ /* Set it back to 0 if there are no waiters: */
+ if (likely(list_empty(&lock->wait_list)))
+ atomic_set(&lock->count, 0);
+
+ spin_unlock_mutex(&lock->wait_lock);
+
+ return prev == 1;
+}
+
+/***
+ * mutex_trylock - try acquire the mutex, without waiting
+ * @lock: the mutex to be acquired
+ *
+ * Try to acquire the mutex atomically. Returns 1 if the mutex
+ * has been acquired successfully, and 0 on contention.
+ *
+ * NOTE: this function follows the spin_trylock() convention, so
+ * it is negated to the down_trylock() return values! Be careful
+ * about this when converting semaphore users to mutexes.
+ *
+ * This function must not be used in interrupt context. The
+ * mutex must be released by the same task that acquired it.
+ */
+int fastcall mutex_trylock(struct mutex *lock)
+{
+ return __mutex_fastpath_trylock(&lock->count,
+ __mutex_trylock_nonatomic);
+}
+
[snip]
+
+#define spin_lock_mutex(lock) spin_lock(lock)
+#define spin_unlock_mutex(lock) spin_unlock(lock)
Is this an interrupt deadlock, or do you not allow interrupt contexts
to even trylock a mutex?
If the former, I would simply use atomic_cmpxchg unconditionally for
the trylock and not use a spinlock fallback function at all.
Alternately, you may 'trylock' the spinlock, but it just seems like
extra code for not much reason (when you can use atomic_cmpxchg).
--
SUSE Labs, Novell Inc.
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